Question

Combustion of a 0.9835g sample of a compound containing only carbon, hydrogen, and oxygen produced 1.900g of CO2 and 1.070g of H2O. Determine the empirical formula of the compound.

Answer 1

CxHyOz+O2-->CO2+H2O
I want to determine each of the element number of moles to calculate the compound empirical formula.
Step 1 Covert their masses of products into moles
12g of C-->44 g of CO2
x g of C-->1.900 g of CO2
x=1.900*12/44=0.518
2 g of H--> 18 g of H2O
y g of H -->1.070 g of H2O
y=1.070*2/18=0.12
Cross multiplication to find the g of both C and H
The amount used of the whole compound is 0.9835 g
g of C + g of H + g Of O=total mass of Compound
0.518+0.12+ g of C=0.9835
0.9835-(0.518+0.12)=0.3455 g of O
Use Number of moles=Mass(g)/Molar Mass(g/mol)
0.518/12=0.0432 moles of C
0.12/1=0.12 moles of H
0.3455/16=0.022 moles of O
Divide by the smallest figure.
0.0432/0.022, 0.12/0.022, 0.022/0.022
2 mole of C 5.5 mole of H 1 mole of O
To get rid of fraction because you cant find a compound with fractions we * by 2
C4H11O2

Answer 2

(1.900 g CO2) / (44.00964 g CO2/mol) x (1 mol C / 1 mol CO2) x (12.01078 g C/mol) =
0.518534 g C
(1.070 g H2O) / (18.01532 g H2O/mol) x (2 mol H / 1 mol H2O) x (1.007947 g H/mol) =
0.119732 g H
Use the Law of Conservation of Mass:
0.9835 g total - 0.518534 g C - 0.119732 g H = 0.345234 g O
Convert to moles:
(0.518534 g C) / (12.01078 g C/mol) = 0.0431724 mol C
(0.119732 g H) / (1.007947 g H/mol) = 0.118788 mol H
(0.345234 g O) / (15.99943 g O/mol) = 0.0215779 mol O
Divide by the smallest number of moles:
(0.0431724 mol C) / 0.0215779 mol = 2.0008
(0.118788 mol H) / 0.0215779 mol = 5.5051
(0.0215779 mol O) / 0.0215779 mol = 1.0000
In order to achieve integer coefficients, multiply by 2, then round to the nearest whole numbers to find the empirical formula:
C4H11O2

Answer 3

Thank you!