The empirical formula of the oxygenated hydrocarbon is CH2O.
To find the empirical formula of the oxygenated hydrocarbonWe need to determine the mole ratios of the elements present in the compound using the given masses and the molar masses of the elements.
First, we need to find the moles of carbon dioxide and water vapor produced in the reaction:
moles of CO2 = 1.76 g / 44.01 g/mol = 0.0400 mol CO2
moles of H2O = 1.08 g / 18.02 g/mol = 0.0599 mol H2O
Next, we need to find the moles of carbon and hydrogen in the compound. We can do this by using the masses of carbon dioxide and water vapor produced, the molar masses of the elements, and the balanced chemical equation for the combustion of hydrocarbons:
CxHyOz + (x + y/4 - z/2) O2 -> x CO2 + (y/2) H2O
From the balanced equation, we can see that the mole ratio of carbon to carbon dioxide is 1:1, and the mole ratio of hydrogen to water vapor is (1/2):(1/2) or 1:1. Therefore:
moles of C = moles of CO2 = 0.0400 mol CO2
moles of H = 2 × moles of H2O = 0.1198 mol H2O
Finally, we can find the mole ratio of carbon to hydrogen in the compound:
mol ratio of C to H = (moles of C) / (moles of H) = 0.0400 mol CO2 / 0.1198 mol H2O ≈ 0.334
To get whole numbers for the mole ratio, we can multiply both sides by a common factor, such as 3:
mol ratio of C to H ≈ 0.334 ≈ 1/3
mol ratio of C to H = 1
mol ratio of O to H = (moles of O) / (moles of H) = (moles of CxHyOz - moles of C - moles of H) / (moles of H) = (0.0400 mol CO2 + 0.0599 mol H2O - 0.0400 mol C - 0.1198 mol H) / 0.1198 mol H = 0.213
To get whole numbers for the mole ratio, we can multiply both sides by a common factor, such as 5:
mol ratio of O to H ≈ 0.213 ≈ 2/9
mol ratio of O to H = 2
Therefore, the empirical formula of the oxygenated hydrocarbon is CH2O.
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Use the E2 mechanism to explain why when I is mixed with sodium ethoxide (NaOCH2CH3) in ethanol, the major product
is III, but when II is mixed with sodium ethoxide in ethanol, the major product is IV.
Two alkenes may be produced when sodium ethoxide is used to treat 2-Bromo-3-methylbutane in ethanol. Saytzeff's rule governs the reaction pathway, leading to the formation of 2 methyl 2 butene as the main product, which is a more substituted alkene.
What is the purpose of sodium ethoxide?Inorganic synthesis uses sodium ethoxide, 21% w/w in ethanol, as a powerful base. It is used in a variety of chemical processes, including esterification, alkoxylation, condensation, and etherification. It participates actively in the Wolf-Kishner reduction, Stobbe reaction, and Claisen condensation.
Because ethoxide ions are Brnsted-Lowry bases and remove hydrogen ions from water molecules to form hydroxide ions, which raise the pH, the solution is extremely alkaline.
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Why C is the correct answer? (the question is:Which of the following compounds are NOT aliphatic hydrocarbons?)
C is the correct option as it contains oxygen along with carbon and hydrogen. Aliphatic hydrocarbons that have had one or more hydrogen atoms .
Aliphatic hydrocarbons that have had one or more hydrogen atoms replaced by a halogen, such as those that have been fluorinated, chlorinated, brominated, or iodized, are known as halogenated aliphatic hydrocarbons.
The group of chemical molecules known as chlorinated aliphatic hydrocarbons is diverse and is characterised by an open-chain structure or a variable number of bonds, which can be single, double, or triple. C is the correct option as it contains oxygen along with carbon and hydrogen.
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How many oxygen atoms are in 225 g 02?
A 225-gram a sample of oxygen contains 84.684375 X 1023 atoms of oxygen.
What do you mean by oxygen, O2?One oxygen atom, designated O. Two molecules of oxygen combine to form the molecule O2. Our bodies have developed to breathe O2, which is the main component of our atmosphere.
Why is oxygen abbreviated O2?Because each molecule is made up of two atoms of oxygen fused together, its formula is written as O2.
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3. 20.0 grams of NaHCO3 decomposes following the balanced reaction below. How many grams of H2O will you form? Show all work. 2 NaHCO3 → H2O + CO2 + Na2CO3
Taking into account the reaction stoichiometry, 2.14 grams of H₂O are formed when 20.0 grams of NaHCO₃ decomposes.
Reaction stoichiometryIn first place, the balanced reaction is:
2 NaHCO₃ → H₂O + CO₂ + Na₂CO₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
NaHCO₃: 2 moleH₂O: 1 moleCO₂: 1 moleNa₂CO₃: 1 moleThe molar mass of the compounds is:
NaHCO₃: 84 g/moleH₂O: 18 g/moleCO₂: 44 g/moleNa₂CO₃: 106 g/moleBy reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
NaHCO₃: 2 moles ×84 g/mole= 168 gramsH₂O: 1 mole ×18 g/mole= 18 gramsCO₂: 1 mole ×44 g/mole= 44 gramsNa₂CO₃: 1 mole ×106 g/mole= 106 gramsMass of H₂O formedThe following rule of three can be applied: if by reaction stoichiometry 168 grams of NaHCO₃ form 18 grams of H₂O, 20 grams of NaHCO₃ form how much mass of H₂O?
mass of H₂O= (20 grams of NaHCO₃× 18 grams of H₂O)÷168 grams of NaHCO₃
mass of H₂O= 2.14 grams
Finally, 2.14 grams of H₂O are formed.
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_C₂H₄+ _ O₂ → _ CO₂ + _ H₂O If you start with 14.5 grams of ethylene (C₂H₄), how many grams of water(H₂O) will be produced?
Answer:
a. SOLUTION:
Step 1: Write the balanced chemical equation.
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
Step 2: Calculate the number of moles of CO₂ formed by each reactant.
• Using C₂H₄
Based on the balanced chemical equation, 1 mole of C₂H₄ is stoichiometrically equivalent to 2 moles of CO₂.
The molar mass of C₂H₄ is 28.054 g/mol.
• Using O₂
Based on the balanced chemical equation, 3 moles of O₂ is stoichiometrically equivalent to 2 mole of CO₂.
The molar mass of O₂ is 31.998 g/mol.
Step 3: Determine the limiting reagent.
Since O₂ produced less amount of CO₂ than C₂H₄, O₂ is the limiting reagent.
Step 4: Determine the mass of CO₂ formed.
Note that the (maximum) mass of a product that can be formed is dictated by the limiting reagent. In this case, we will start at the number of moles of CO₂ formed from the limiting reagent (O₂) which is equal to 0.11022 mol.
The molar mass of CO₂ is 44.009 g.
Hence, 4.85 g of CO₂ can be formed.
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b. ANSWER:
The LR is O₂ and the ER is C₂H₄.
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c. SOLUTION:
The theoretical yield of the reaction is 4.85 g.
Hence, the percent yield of the reaction is 87.6%.
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Explanation:
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the H2 produced in a chemical reaction is collected through water in a eudiometer. If the pressure in the eudiometer is 760.0 torr and the vapor pressure of water under the experimetal conditions is 20.9 torr what is the pressure (torr) of the H2 gas?
The pressure of the H2 gas is 739.1 torr.
What is Pressure?
The SI unit of pressure is Pascal (Pa), which is defined as one Newton per square meter (N/m^2). Other common units of pressure include atmospheres (atm), millimeters of mercury (mmHg), torr, and pounds per square inch (psi). Pressure is a fundamental concept in physics, chemistry, and engineering, and it plays a crucial role in various natural and man-made processes, such as fluid mechanics, thermodynamics, and weather patterns.
To find the pressure of the H2 gas, we need to subtract the vapor pressure of water from the total pressure inside the eudiometer.
Given:
Pressure of eudiometer (total pressure) = 760.0 torr
Vapor pressure of water = 20.9 torr
Pressure of H2 gas = Pressure of eudiometer - Vapor pressure of water
Pressure of H2 gas = 760.0 torr - 20.9 torr
Pressure of H2 gas = 739.1 torr
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