Question

1. Vanillin, a flavoring agent, is made up of carbon, hydrogen, and oxygen atoms. When a sample of vanillin weighing 2.500g burns in oxygen, 5.79 g of carbon dioxide and 1. 18 g of water are obtained. What is the empirical formula of this compound? 2- A 0.250 g sample of hydrocarbon undergoes complete combustion to produce 0.845 g of CO2 and 0.173 g of H2O. What is the empirical formula of this compound?

          1. Vanillin, a flavoring agent, is made up of carbon, hydrogen, and oxygen atoms. When a sample of vanillin weighing 2.500g burns in oxygen, 5.79 g of carbon dioxide and 1. 18 g of water are obtained. What is the empirical formula of this compound?

2- A 0.250 g sample of hydrocarbon undergoes complete combustion to produce 0.845 g of CO2 and 0.173 g of H2O. What is the empirical formula of this compound?

1. Vanillin, a flavoring agent, is made up of carbon, hydrogen, and oxygen atoms. When a sample of vanillin weighing 2.500g burns in oxygen, 5.79 g of carbon dioxide and 1. 18 g of water are obtained. What is the empirical formula of this compound?

2- A 0.250 g sample of hydrocarbon undergoes complete combustion to produce 0.845 g of CO2 and 0.173 g of H2O. What is the empirical formula of this compound?

Added by Joseph C.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Hitendra Singh

11/13/2021

Step 1

- Given: 5.79 g of CO₂ - Molar mass of CO₂ = 44.01 g/mol - Moles of CO₂ = 5.79 g / 44.01 g/mol = 0.1316 mol - Each mole of CO₂ contains 1 mole of carbon, so moles of carbon = 0.1316 mol Show more…

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1. Vanillin, a flavoring agent, is made up of carbon, hydrogen, and oxygen atoms. When a sample of vanillin weighing 2.500g burns in oxygen, 5.79 g of carbon dioxide and 1.18 g of water are obtained. What is the empirical formula of this compound? 2- A 0.250 g sample of hydrocarbon undergoes complete combustion to produce 0.845 g of CO2 and 0.173 g of H2O. What is the empirical formula of this compound?