Question

A compound contains only carbon, hydrogen, and oxygen. Combustion of 91.88 g of the compound yields 134.7 g of CO2 and 55.13 g of H2O. The molar mass of the compound is 180.156 g/mol.

1. Calculate the grams of carbon (C) in 91.88 g of the compound: grams

2. Calculate the grams of hydrogen (H) in 91.88 g of the compound. grams

3. Calculate the grams of oxygen (O) in 91.88 g of the compound. grams

1. the moles of carbon (C) in 91.88 g of the compound: moles

2. The moles of hydrogen (H) in 91.88 g of the compound:

3. the moles of oxygen (O) in 91.88 g of the compound:

Divide each mole quantity that you determined in the previous question by the smallest number of moles to determine the correct empirical formula. Enter the correct subscript for each atom using the smallest whole number. Enter a 1 if that is the smallest whole number, don't leave the box empty. C H O Tries 0/99 Now determine the molecular formula. Remember that the molar mass of the compound is 180.156 g/mol. Enter the correct subscript for each atom using the smallest whole number. Enter a 1 if that is the smallest whole number, don't leave the box empty. C= H= O=

Answer 1

moles of C = moles of CO2
= mass of CO2/molar mass of CO2
= 134.7/44
=3.0614 mol

masss of C = moles of C* molar mass of C
= 3.0614*12
=36.7364 g

moles of H = 2*moles of H2O
= 2*mass of H2O/molar mass of H2O
= 2*55.13/18
=6.1256 mol

masss of H = moles of H* molar mass of H
= 6.1256*1
=6.1256 g

mass of O = 91.88 - 36.7364 - 6.1256 =49.018 g
moles of O =mass of O / molar mass of O
= 49.018/16
= 3.0626 mol

So we have:
C : 3.0614 mol
H : 6.1256 mol
O : 3.0626 mol

divide by smallest
C : 3.0614/3.0614 = 1
H : 6.1256/3.0614 = 2
O : 3.0626/3.0614 = 1

Answer:
C = 1
H = 2
O = 1