A compound contains only carbon, hydrogen, and oxygen. Combustion of 16.51 g of
ID: 538942 • Letter: A
Question
A compound contains only carbon, hydrogen, and oxygen. Combustion of 16.51 g of the compound yields 24.21 g of CO2and 9.909 g of H2O.
The molar mass of the compound is 150.13 g/mol.
1. Calculate the grams of carbon (C) in 16.51 g of the compound:
2. Calculate the grams of hydrogen (H) in 16.51 g of the compound:
3. Calculate the grams of oxygen (O) in 16.51 g of the compound:
Based on your previous answers, calculate
1. the moles of carbon (C) in 16.51 g of the compound:
2. the moles of hydrogen (H) in 16.51 g of the compound:
3. the moles of oxygen (O) in 16.51 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_
Now determine the molecular formula. Remember that the molar mass of the compound is 150.13 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_
Explanation / Answer
CO2 mass = 24.21 g
CO2 moles = 24.21 / 44
= 0.550
moles of C = 0.550
moles of C = 0.550 x 12
mass of C =0.550 x 12
mass of C = 6.60 g
moles of H2O = 9.909 / 18 = 0.5505
moles of H = 2 x 0.5505 = 1.101
moles of H = 1.101
mass of H = 1.101 x 1
mass of H = 1.101 g
mass of O = 16.51 - (6.60 + 1.101)
= 8.809 g
mass of O = 8.809 g
moles of O = 8.809 / 16 = 0.551
moles of O = 0.551
moles ratio = 0.550 : 1.101 : 0.551
= 1 : 2 : 1
empirical formula = CH2O
mass of empirical formula = 12 +2 + 16 = 30
n = 150 / 30 = 5
molecular formula = 5 x CH2O
= C5H10O5
molecular formula = C5H10O5
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