the molecule has the empirical formula of c3h4o2 the molectular massis 144g/mol the molecular formula for this molecule is

The percent of the carbon in the octane that is not fully oxidized is 3.8%. This is due to incomplete combustion of the octane, resulting in the production of carbon monoxide (CO) and other pollutants.

The combustion of octane is a highly exothermic reaction. For every three oxygen molecules (O2) that are consumed, eight carbon dioxide molecules (CO2) and eighteen water molecules (H2O) are produced. The theoretical energy output of the complete combustion of octane is 3.7 megajoules per kilometer (MJ/km).

However, due to incomplete combustion, not all of the octane is fully oxidized. Incomplete combustion of octane results in the formation of carbon monoxide (CO) and other pollutants. In a gasoline-powered vehicle, 1.7 grams per kilometer (g/km) of CO is often emitted due to incomplete combustion.

To determine the percent of the carbon in the octane that is not fully oxidized, we can use the following equation:

Percent Carbon Not Fully Oxidized = (CO emission rate (g/km) / Carbon content of octane (g/km)) x 100

In our example, the CO emission rate is 1.7 g/km and the carbon content of octane is 44.1 g/km. Thus, the percent of the carbon in the octane that is not fully oxidized is 3.8%. This means that 3.8% of the carbon in the octane is not being completely oxidized in the combustion process, resulting in the formation of CO and other pollutants.

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The pH will fall below 7.00. Ammonia makes up 0.050 moles. In water, a polar solvent, an ionic solute dissolves because the polar water molecules draw the ions into solution, where they hydrate.

This means that equal amounts of moles of sodium hydroxide and hydrochloric acid are needed for a full neutralization to produce a neutral solution, or a solution with a pH of 7 at ambient temperature. As a result, a substance's molar mass, also known as its molecular mass or Mm, is equal to its atomic or molecular weight stated in grams. Determine the pH of a solution that contains 0.20 M ammonium chloride and 0.10 M aqueous ammonia. As you are aware, sodium hydroxide and hydrochloric acid have a mole ratio of 1:1 for neutralization.

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The age of that rock will be A. 2,000 million years.

The half-life period can be mathematically represented as follows,

t1/2 = 0.693/ Decay Constant

Half-life and the radioactive decay rate constant λ are inversely proportional which means the shorter the half-life, the larger λ and the faster the decay.

Decay Constant = 0.693/1000 * 106 years = 6.93*10-10 yr-1

We know that the radioactive decay equation as,

N = N0 e- (Decay Constant * Time)

N/N0 = 1/3 = e(6.93*10-10 yr-1 * time)

Time = 1600 * 106 years.

Time=2,000 million years

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Complete question:

The Half-Life Of A Radioactive Isotope Is 1,000 Million Years. Scientists Testing A Rock Sample Discover That The Sample Contains Three Times As Many Daughter Atoms As Parent Isotopes. What Is The Age Of The Rock? A. 2,000 Million B. 1,000 Million C. 500 Million D. 250 Million

Answer:

Explanation:

To solve this problem, we need to use the equation Q = mcΔT, where Q is the heat transfer, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

First, we need to determine the mass of water in the pot. Let's assume that the pot has a volume of 1 liter, or 1000 mL. Water has a density of 1 g/mL, so the mass of the water is 1000 g.

Next, we need to determine the change in temperature of the water. We know that the heat transfer is 640 KJ, and the specific heat capacity of water is 4.184 J/g°C. So, the change in temperature of the water is 640,000 J / (4.184 J/g°C * 1000 g) = 152.3°C.

Now, we can use the equation Q = mcΔT to solve for the mass of propane needed. We know that Q = 640,000 J, m = 1000 g, c = 4.184 J/g°C, and ΔT = 152.3°C. Plugging these values into the equation, we get 640,000 J = (1000 g)(4.184 J/g°C)(152.3°C). Solving for the mass of propane, we get m = 640,000 J / (4.184 J/g°C * 152.3°C) = 30.8 g.

So, the amount of propane needed to transfer 640 KJ of heat into the pot of water is 30.8 g.

Answer:

approximately 22 grams of propane to transfer 640 KJ of heat into a pot of water.

Explanation:

To determine the mass of propane needed to transfer 640 KJ of heat into a pot of water, you can use the following equation:

mass (g) = (heat transfer (J) / molar enthalpy of combustion (J/mol)) x molar mass (g/mol)

First, you need to convert the heat transfer value from KJ to J. To do this, multiply the value in KJ by 1000: 640 KJ * 1000 J/KJ = 640000 J

Next, you need to convert the molar enthalpy of combustion value from KJ/mol to J/mol. To do this, multiply the value in KJ/mol by 1000: -2220 KJ/mol * 1000 J/KJ = -2220000 J/mol

Finally, you can plug these values into the equation above to calculate the mass of propane needed:

[tex]mass (g) = (640000 J / -2220000 J/mol) x 44 g/mol = 22 g[/tex]

So, you would need approximately 22 grams of propane to transfer 640 KJ of heat into a pot of water.

A compound forms a dense yellow precipitate when treated with iodine and sodium hydroxide. the compound must be: Acetophenone methyl ketone.

Sodium hydroxide is every now and then known as caustic soda or lye. it's by far a common ingredient in cleaners and soaps. At room temperature, sodium hydroxide is white, odorless strong. Liquid sodium hydroxide is colorless and has no odor. it could react violently with strong acids and with water.

Sodium hydroxide, also known as lye and caustic soda, is an inorganic compound with the formula NaOH. it is a white stable ionic compound that includes sodium cations Na⁺ and hydroxide anions OH⁻.

Manufacturers may additionally use sodium hydroxide to provide soaps, rayon, paper, products that explode, dyes, and petroleum merchandise. different obligations which can use sodium hydroxide include processing cotton cloth, steel cleaning and processing, oxide coating, electroplating, and electrolytic extraction.

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Answer:

Yellow, Purple

Explanation:

5. The law of conservation of mass essentially states that the total 'mass' that you started with equals to the total 'mass' at the end. Take the burning of a tree, for example. Although the tree loses mass, the mass of carbon dioxide and other molecules increases proportionally to the rate of tree mass loss. Abiding by this law in a chemical reaction, we should expect the amount of shapes on each side to be equal, although they may not necessarily be in the same form. This is only shown in the 'Yellow' square.

6. We want the same amount of stuff on the left as we do on the right. There is 2 Nitrogen and 2 Hydrogen on the left. There is 1 Nitrogen and 3 Hydrogen on the right. To balance these amounts out, we'd need to multiply a few things:

1 Nitrogen by 2 on the right to make things even, meaning that now there is 2 Nitrogen and 6 Hydrogen on the right. To balance this out on the left, we need to multiply the 2 Hydrogen by 3 to yield 6 Hydrogen.

As such, the new equation is N2+ 3H2 --> 2NH3.

Do we have the same amount of stuff on the left as we do on the right? Yes! So the equation is balanced.

The conditions for a standard electrochemical cell. select one or more : pressure of 1 atm temperature of 298 k solution concentrations of 1 M.

The electrochemical cell is the cell that is capable of generating the electrical energy from the chemical reactions or by the use of the electrical energy to cause the chemical reaction. The conditions for a standard electrochemical cell. select one or more : pressure of 1 atm temperature of 298 k solution concentrations of 1 M.  

There are the two types of the electrochemical cells is as follows : the galvanic called the electrolytic cells. the galvanic cell is also called as the voltaic cell.

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True, This occurs because as nuclear charge increases, losing electrons from an atom becomes more challenging.

The atom radius of an atom is the distance between its nucleus's centre and its outermost shell. As we transition from one era to another, the atomic size of a group grows as a result of the addition of shells. Over time, the number of shells stays constant as the nuclear charge rises, but the atomic size shrinks. As a result, electrons in the outermost shell are drawn toward the nucleus, causing the size to decrease.

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The empirical formula is obtained by compositons: Cl2Zn for the empircal formula of a compound, given that the compound is found to be 47.9% zinc and 52.1% chlorine by mass

Calculate the molar mass of each element: Cl=35.453, Zn=65.409

Cl=1.469551236848, Zn=0.73231512482992 when expressed in moles.

Identify the least mole value: 0.73231512482992

Multiplying all elements by their lowest value Cl=2.0067197672451, Zn=1

rounding to the nearest whole number Cl=2, Zn=1

Cl[tex]_{2}[/tex]Zn for the empircal formula

The empirical formula, which is defined as the ratio of compositons of the least whole number of the components included in the formula, is the simplest formula for a compound compositons. The simplest formula is another name for it.

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An alloy is made of two or more different metals.These metals are of different sizes. This cause a distortion in the system and leads to a slide of of atoms when a force is applied and make it brittle.

Alloys are mixed metals. They are widely used in constructions, in electronic devices and in other industries due to their potential application superior over the pure metals.

There are atoms of various sizes in an alloy. The layers of atoms in pure metal are distorted by the smaller or larger atoms. This indicates that more force is needed to get the layers to slide over one another. Compared to pure metal, the alloy is tougher and more durable.

The irregularity of the atoms that make up high-entropy alloys is what gives is their immense strength. When the alloy is deformed, this disorder makes it difficult for the alloy's dislocation flaws to migrate through its crystal structure. But under adequate pressure, this also makes the alloy highly brittle.

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Two isotopes of the same element only have these differences: atomic mass

Isotopes are forms of an element that have the same number of protons and electrons but differing numbers of neutrons. The various isotopes of an element have varying weights depending on the amount of neutrons present in each isotope.

Chemically speaking, an element's isotopes are nearly or completely identical. The chemical behaviours of several isotopes are quite similar. But each isotope has a unique set of physical characteristics, including as mass, melting or boiling temperature, density, and freezing point. Any isotope's physical characteristics are mostly influenced by its mass.

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Answer:

The IUPAC names for Ascorbic Acid, Fructose, and Glucose are L-ascorbic acid, D-fructose, and D-glucose, respectively.

Explanation:

IUPAC (International Union of Pure and Applied Chemistry) names are standardized names used to identify chemical compounds. These names are based on the structure of the compound, and they provide a unique and unambiguous way to identify each compound.

The IUPAC name for Ascorbic Acid is L-ascorbic acid. Ascorbic acid is a compound with the chemical formula C6H8O6. It is a white, crystalline solid that is soluble in water. The "L" in the name indicates that the compound has a specific configuration of atoms, which is called the L-configuration.

The IUPAC name for Fructose is D-fructose. Fructose is a simple sugar with the chemical formula C6H12O6. It is a sweet-tasting substance that is commonly found in fruits and honey. The "D" in the name indicates that the compound has a specific configuration of atoms, which is called the D-configuration.

The IUPAC name for Glucose is D-glucose. Glucose is a simple sugar with the chemical formula C6H12O6. It is a sweet-tasting substance that is commonly found in plants and is an important source of energy for living organisms. The "D" in the name indicates that the compound has a specific configuration of atoms, which is called the D-configuration.

Answer:

The IUPAC name for ascorbic acid is 2-(1,2-dihydroxyethyl)-4,5-dihydroxyfuran-3-one.

The IUPAC name for fructose is (2R,3S,4R,5R)-2-(hydroxymethyl)-3,4,5-trihydroxytetrahydrofuran-3-ulose.

The IUPAC name for glucose is (2R,3R,4S,5S)-2-(hydroxymethyl)-3,4,5-trihydroxytetrahydrofuran-3-ulose.

If an isotope with a mass number of 124 in a sample of 5.63 g decays at a rate of 0.300 ci, its half-life is 5.00*10^4years, as is described below.

Half-life is the length of time it takes for half of an unstable nucleus to go through its decay process. A radioactive element's half life decay time varies depending on the element. The entire number of protons and neutrons (collectively known as nucleons), which make up an atomic nucleus, is known as the mass number (symbol A, from the German word Atomgewicht [atomic weight]). It is roughly equivalent to the atomic mass of the atom, also known as its isotopic mass, given in atomic mass units.The response to the posed query is included below.

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Answer: 29.4L

Explanation:

n = m/mm

36.8g/28g/mol = 1.31mol

At STP, the molar volume is equal to 22.4L/mol, therefore, the volume that will be occupied by 36.8g carbon monoxide is:

Multiply 22.4 and 1.31,

Answer = 29.4L

If 4 moles of zinc hydroxide react, The reaction consumes2 moles of sulfuric acid The reaction produces _2_ moles of zinc sulfate and _2__ moles of water

Zinc sulfate is the inorganic compound with the components ZnSO4 and traditionally recognised as "white vitriol".

It is at the World Health Organization's List of Essential Medicines, a listing of the maximum vital remedy wanted in a simple fitness system.neutralization reactionThe given response is a neutralization response. Zinc oxide (base) reacts with sulfuric acid (acid) to shape zinc sulfate (salt) and water.

When sulfuric acid reacts with zinc hydroxide, zinc sulfate and water are produced. The balanced equation for this response is: H2SO4(aq)+Zn(OH)2(s)=ZnSO4(aq)+2H2O(1) If three moles of zinc hydroxide react, The response consumes _____ moles of sulfuric acid.

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Yes it could, but you'd have to set up the process very carefully.

I see two major challenges right away:

1).  Displacement of water would not be a wise method, since rock salt

is soluble (dissolves) in water.  So as soon as you start lowering it into

your graduated cylinder full of water, its volume would immediately start

to decrease.  If you lowered it slowly enough, you might even measure

a volume close to zero, and when you pulled the string back out of the

water, there might be nothing left on the end of it.

So you would have to choose some other fluid besides water ... one in

which rock salt doesn't dissolve.  I don't know right now what that could

be.  You'd have to shop around and find one.

2).  Whatever fluid you did choose, it would also have to be less dense

than rock salt.  If it's more dense, then the rock salt just floats in it, and

never goes all the way under.  If that happens, then you have a tough

time measuring the total volume of the lump.

So the displacement method could perhaps be used, in principle, but

it would not be easy.

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a) The number of moles of KCl formed is 0.51 moles.

b) The number of molecules of KCl is 1.38 * 10^22 molecules.

c) The mass of the potassium chlorate 8.94 g.

We know that we can be able to use the stoichiometry of the reaction to be able to find the amount of the KCl that is formed. The first step is to write down the reaction equation as it has been shown;

[tex]2KClO_{3} ----- > 2KCl + 3O_{2}[/tex].

We can see that we can be able to apply the principles of stoichiometry as stated earlier.

a) Number of moles of the potassium chlorate = 62 g/122.55 g/mol

= 0.51 moles

Then we have the number of the moles of the KCl as 0.51 moles.

b) Number of moles of the potassium chlorate =  2.85 g/122.55 g/mol

= 0.023 moles

If 2 moles of potassium chlorate produces 2 moles of KCl then 0.023 moles produces 0.023 moles of KCl.

Number of molecules of KCl = 0.023 moles * 6.02 * 10^23

= 1.38 * 10^22 molecules

c) If the number of moles of oxygen is; 3.54 g/32 g/mol = 0.11 moles

If 2 moles of potassium chlorate produces 3 moles of oxygen

x moles of potassium chlorate produces  0.11 moles

x = 2 * 0.11/3

= 0.073 moles

Mass of the potassium chlorate =  0.073 moles * 122.55 g/mol

= 8.94 g

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Answer:

In this experiment, the variable is the substance that was placed in the water. The student performed the experiment with four different substances (sodium chloride, vegetable oil, sugar, and glycerin), and she measured the effect of each substance on the water. All other factors, such as the volume of the water and the time that she stirred each beaker, were kept constant. Therefore, the correct answer is option C: "the substances that were placed in the water."

The vapor pressure of the glucose solution at 25°C is 20.16 mmHg.

To answer this question, we need to use Raoult's Law. Raoult's Law states that the vapor pressure of a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution.

In this case, the solvent is water and the solute is glucose. We can use the given information to calculate the mole fraction of water in the solution:

Mole fraction of water = (72.0 g H2O) / (72.0 g H2O + 18.0 g C6H12O6) = 0.8

We can then use Raoult's Law to calculate the vapor pressure of the glucose solution:

Vapor pressure of glucose solution = (Vapor pressure of pure water) * (Mole fraction of water in solution)

= (25.2 mmHg) * (0.8)

= 20.16 mmHg

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The freezing point of the solution will be - 0.952 °C can be calculated by the molality of the solution.

Molality of the solution can be calculated as follows by :

Molality= Number of the moles of the solute / weight of the solvent in Kg

= 27.3 ×100/ 83.0 × 46.07

= 0.7m

Since the molar mass of the ethanol is 46.07g/mol

So by the formula,

The molal freezing point depression constant KF and the solute's molality, m, are used to calculate the freezing point depression, which is given by T = KFm. Rearranging results in the formula: mol solute = (m) x (kg solvent).

ΔTf​ = Kf  ×Molality

=1.86×0.7

= 0.952

Thus, freezing point = 0−0.952

                                 =−0.952

Hence the freezing point of the resulting solution is calculated .

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Answer:

an article published by a university or a government agency

and

an article published in 2015 and updated a week ago

Explanation:

Both of these meet the criteria for a credible source. Cites with Advertisements, blogposts, and anonymous cites do not

There are approximately 0.000486 moles of hydrogen present in the sample.

To find the number of moles of hydrogen present in the sample, you need to use the ideal gas law equation, which is PV = nRT where P is the pressure of the gas (in atm), V is the volume of the gas (in L), n is the number of moles of gas, R is the ideal gas constant, and T is the temperature of the gas (in K). First, you need to convert the pressure from torr to atm. You can do this by dividing the pressure in torr by 760, since 1 atm is equal to 760 torr. In this case, the pressure in atm would be: P = 764.5 torr / 760 = 1.0059 atm Next, you need to convert the volume from ml to L. You can do this by dividing the volume in ml by 1000, since 1 L is equal to 1000 ml. In this case, the volume in L would be: V = 30.42 ml / 1000 = 0.03042 L Finally, you need to convert the temperature from degrees Celsius to Kelvin. You can do this by adding 273.15 to the temperature in Celsius. In this case, the temperature in K would be:T = 25.58 degrees Celsius + 273.15 = 298.73 K Now that you have all the necessary values, you can plug them into the ideal gas law equation to solve for the number of moles of hydrogen: n = (PV) / (RT) = (1.0059 atm * 0.03042 L) / (8.31 J/mol*K * 298.73 K) = 4.86 x 10^-4 mol Therefore, there are approximately 0.000486 moles of hydrogen present in the sample

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37.4 ml of volume of 0.209 M KMnO4 solution is needed to prepare 487. ml of 0.0158 m KMnO4.

we need to use the concept of dilution. Dilution is the process of preparing a lower concentration solution by adding more solvent (in this case, water) to a concentrated solution. The concentration of a dilute solution can be calculated using the following formula:

C1 * V1 = C2 * V2

where C1 is the concentration of the original (concentrated) solution, V1 is the volume of the original solution, C2 is the concentration of the dilute solution, and V2 is the volume of the dilute solution.

In this case, we are given that the concentration of the original solution is 0.209 M and the concentration of the dilute solution is 0.0158 M. We are also given that the volume of the dilute solution is 487 mL. We can use these values to solve for the volume of the original solution as follows:

V1 = (C2 * V2) / C1

= (0.0158 M * 487 mL) / 0.209 M

= 37.4 mL

Therefore, the volume of the 0.209 M KMnO4 solution needed to prepare 487 mL of 0.0158 M KMnO4 is 37.4 mL.

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Is the highest force between molecules Option a) water. hydrogen bonds are responsible for holding nucleotide bases together in DNA and RNA.

When a hydrogen atom is bound to either an oxygen, nitrogen, or fluorine atom, a peculiar type of dipole-dipole interaction takes place. In another molecule, the partially negative end of the oxygen, nitrogen, or fluorine is drawn to the partially positive end of the hydrogen atom.

A significant amount of energy is needed to break a hydrogen bond since it is a rather powerful force of attraction between molecules. This explains why substances like hydrogen fluoride (H, F) and water, H2O, have melting and boiling temperatures that are incredibly high. For instance, hydrogen bonds hold the nucleotide bases in DNA and RNA together, which is a crucial function of hydrogen bonds in biology.

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We need 5.36 hours of constant current supply through the electrolytic cell containing aqueous copper ions.

What is an electrolytic cell ?

Among the different types of using current to power an electrolytic cell is the source of energy for any prospective growth across the cell where the redox chemical reaction takes place

Given:

Current (I) = 50 A

Quantity of electricity ( Q) = I x t

The half cell reaction can written as

Cu [tex]Cu^{2+} + 2e^{-1}[/tex] → [tex]Cu[/tex]

1 mole of copper is deposited by

9500 x 2 = 193000 coloumbs

1 mole of copper = 193000 coloumb

5 mole of copper = x

X = [tex]\frac{193000}{1}[/tex] x 5

= 965000C

putting the value of Q and I

965000 = 50 x t

1 hour = 3600 sec

t = 5.36 hours

thus we need 5.36 hours of constant current supply through the electrolytic cell.

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There are 2.66 x 10⁹ years old rocks.

Solution:

Since the formation rock U-238 gradually converts to Pb-206.

The half-life for the conversion of U-238 to Pb-206 is, t1/2 = 4.50 x 109 years

Now t1/2 = 0.693 / k = 4.50 x 10⁹ years

= k = 0.693 / 4.50 x 109 years  = 1.54 x 10-10 year-1

Given the mass of Pb-206 formed = 0.438

Hence moles of Pb-206 formed = mass / molecular mass = 0.438 / 206 = 0.0021262 mol Pb

The radioactive disintegration reaction is

U-238  ---- > Pb-206

1 mol ---------1 mol

1 moles of U-238 forms 1 mol of Pb-206.

Hence 0.0021262 mol Pb that would be formed by the moles of U-238  = 0.0021262 mol U-238

Hence mass of U-238 = (0.0021262 mol U-238) x (238 g/mol U-238) = 0.506 g U-238

Hence initial amount of U-238, N0 = 1.506 g

Current amount of U-238, Nt = 1.00 g

Now applying the integrated equation for first-order reaction

ln(N0/Nt) = kt

= t = (1/k) x ln(N0/Nt) = (1 / 1.54 x 10-10 year-1) x ln(1.506 g / 1.00 g) = 2.66 x 10⁹ years.

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To solve such this we must know the concept of acid base reaction. Therefore, in below given ways salt can be formed easily. Salt is formed by reaction between acid and base.

Chemical reaction is a process in which two or more than two molecules collide in right orientation and energy to form a new chemical compound. The mass of the overall reaction should be conserved. There are so many types of chemical reaction reaction like combination reaction, double displacement reaction, acid base reaction.

In the following ways salts can be made using nonmetals, oxides and bases.

oxides of non metal + water [tex]\rightarrow[/tex] acid

acid +base [tex]\rightarrow[/tex] salt

Therefore, in above given ways salt can be formed easily.

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An unsusual dining experience I had was at an eatery that provided a live band as well as had wel dressed security personnel who danced to wecome their customers.

The restauarant stood out for me because of the quality of the food and the entertainment. Hence, I decised to alway visit the restaurant anytime I visist the area.

The additional tasks include:

Diners are individuals who are eating out in a restaurant or eatery.

Diners are usually out to get the best dining experience.

In recent times,  Diners are not only interested in a good meal these days; they often want to enjoy entertainment as well.

Hence, eateries or restuarants are now seeking out new ways to provide entertainment to their customers.

They do so by providing the best of meals in great environments as well providing an accompanying entertainment.

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The Czochralski method is the process most commonly used to grow single-crystal ingots of silicon for semiconductor processing.

The Czochralski method, also known as the Czochralski process or Czochralski technique, is a method of crystal growth employed to obtain single crystals of semiconductors such as silicon, germanium, and gallium arsenide, metals such as palladium, silver, platinum, gold), salts and synthetic gemstones.

The most important application of this method may be the growth of large cylindrical ingots, or boules, of single-crystal silicon used in the electronics industry to make semiconductor devices like integrated circuits.

Czochralski's method is not limited to the production of metal or metalloid crystals. It is also employed in the production of very high-purity crystals of salts, also used in particle physics experiments, with tight controls on confounding metal ions and water absorbed during production.

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The empirical formula for the given compound is Al₂Se₃.

Step 1: Write the data given

Mass of the sample of Aluminium = 7.300 grams

Mass of the definite compound = 39.35 grams

Molar mass of the Aluminium = 26.98 g/mol

Molar mass of the Selenium = 78.96 g/mol

Step 2: Calculate the mass of selenium

Mass of selenium = mass of compound - mass of aluminium

Mass of selenium = 39.35 - 7.3 = 32.05 grams

Step 3: Calculate the number moles of Al

Moles Al = Mass Al/ Molar mass Al

Moles Al = 7.300 grams / 26.98 g/mol

Moles Al = 0.2706 moles

Step 4: Calculate the number moles of Se

Moles Se = Mass Se / Molar mass Se

Moles Se =32.05 g / 78.96 g/mol

Moles Se = 0.4059 moles

Step 5: Divide through the smallest amount of moles

Aluminium: 0.2706 / 0.2706 = 1

Selenium: 0.4059/0.2706 = 1.5

This means for each moles of aluminium, we have 1.5 moles of selenium.

For each 2 moles of aluminium, we have 3 moles of selenium

The empirical formula is Al₂Se₃  

Hence, the compound is aluminium(III) selenide.

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