A 14.1 mg sample of a hydrocarbon was burned in
air. The products were 38.8 mg of CO2 and 31.7 mg of water. what
is the emppirical formula of the hydrocarbon.
Note I am using mmols (it is just moles but
smaller)
38.8mg
of CO2 x
1 mmol CO2
=0.882mmol CO2x
1 mmol C
=0.882mmol C
44.0mg CO2
1 mmol CO2
31.7mg H2O
x
1 mmol H2O
=1.76 mmol H2Ox
2 mmol H
= 3.5mmol H
18.0mg H2O
1 mmol H2O
Set up a mole ratio by dividing by
the smaller of the two
0.882mmol C
/ 0.882mmol = 1
Empirical Formula =CH4
3.5mmol H
/ 0.882mmol = 4
A 0.2417g sample of a compound composed of
C,H,O,Cl only, is burned in oxygen yielding 0.4964g of CO2
and 0.0846g of H2O. A separate 0.1696g sample of the
compound is fused with sodium metal, the products dissolved in
water and the chloride quantitatively precipitated with AgNO3
to yield 0.1891g of AgCl. What is the simplest empirical formula
for the compound.
Sample
Convert to moles
Moles of Compounds
Moles of each element
0.4964g
of CO2 x
1 mol CO2
=0.01128mol CO2x
1 mol C
=0.01128mol C
44.0g CO2
1 mol CO2
0.0846g
H2O x
1 mol H2O
=0.00470mol H2Ox
2 mol H
= 0.0094mol H
18.0g H2O
1 mol H2O
0.1891g
AgCl x
1 mol AgCl
=0.001320molAgClx
1 mol Cl
=0.001320mol Cl (from
2nd sample)
143.3g AgCl
1 mol AgCl
Set up a proportion for the moles of silver
0.001320mols Cl
=
X mol Cl
X=0.00188moles Cl (in first sample)
0.1696g
0.2417g
Find the grams of each element
0.01128mol
C x
12.01 g C
=0.1355g C
1 mol C
Sample MassCHClO
0.2417g
0.0094mol H
x
1.00 g H
=0.0667 g Cl
CHCl mass
-0.2117g CHCl
1 mol H
O mass
0.0300g O
0.00188mol Cl x
35.5 g Cl
=0.0094g H
1 mol Cl
TOTAL
0.2117g CHCl
Find the moles of O 0.0300g O x
1 mol O
=0.00187 mol O
16.0g O
Finally find the mole ratio by dividing by the
smallest quantity
