Question

To use combustion analysis data to determine an empirical formula A molecular formula expresses the number of each kind of atom in a molecule. For example, the molecular formula for propene, C3H6, indicates three carbon atoms and six hydrogen atoms per molecule. This also means that one mole of propene contains three moles of carbon and six moles of hydrogen. An empirical formula expresses the mole ratio of the elements. The empirical formula for propene is CH2, indicating twice as much hydrogen as carbon. When analyzing unknown compounds in a lab, it is often possible to identify the mole ratios, and thus the empirical formula, but not the molecular formula. Notice that the molecular mass of propene, 3(12)+6(1)=42amu, is a multiple of the empirical formula mass, 1(12)+2(1)=14amu . An unknown compound contains only carbon, hydrogen, and oxygen (CxHyOz). Combustion of 3.00 g of this compound produced 4.40 g of carbon dioxide and 1.80 g of water.How many moles of carbon, C, were in the original sample?How many moles of hydrogen, H, were in the original sample?

Answer 1

Answer:

0,1 mol of carbon and 0.2 moles of hydrogen were in the original sample.

Explanation:

The combustion reaction of the compound is

$C_{x} H_{y} O_{z} +O_{2} -> CO_{2} + H_{2}O$

And the molar mass of CO2 and H2O are 44g/mol and 18g/mol respectively

All the mass of carbon producing the CO2 must come from the carbon of the unknown compound, thus it is necessary to calculate the moles of carbon in 4.4g of CO2

$\frac{4.4gx12g}{44gx12g/mol} = 0.1 mol$ of Carbon

Also, all the mass of hydrogen producing the H2O must come from the hydrogen of the unknown compound, hence it is necessary to calculate the moles of hydrogen in 1.8g of water

$\frac{1.8gx2g}{18gx1g/mol} = 0.2 mol$ of Hydrogen