A compound contains only carbon, hydrogen, and oxygen. Combustion of 65.76 g of

Question

A compound contains only carbon, hydrogen, and oxygen. Combustion of 65.76 g of the compound yields 96.38 g of CO2 and 39.46 g of H2O.
The molar mass of the compound is 90.078 g/mol.

1. Calculate the grams of carbon (C) in 65.76 g of the compound:
2. Calculate the grams of hydrogen (H) in 65.76 g of the compound.
3. Calculate the grams of oxygen (O) in 65.76 g of the compound.

Based on your previous answers, calculate

1. the moles of carbon (C) in 65.76 g of the compound:

2. the moles of hydrogen (H) in 65.76 g of the compound:
3. the moles of oxygen (O) in 65.76 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.

_C _H _O

Now determine the molecular formula. Remember that the molar mass of the compound is 90.078 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

Answer –

We are given, mass of compound = 65.76 g

Mass of CO2 = 96.38 g , mas of H2O = 39.46 g

First we need to calculate moles of CO2 and H2O from the given moles

Moles of CO2 = 96.38 g / 44.0 g.mol-1 = 2.19 mole

Moles of H2O = 39.46 g / 18.015 g.mol-1 = 2.19 moles

Moles of C from the moles of CO2

1 moles of CO2 = 1 moles of C

So, 2.19 moles of CO2 = ?

= 2.19 moles of C

Moles of H from moles of H2O

1 moles of H2O = 2 moles of H

So, 2.19 moles of H2O = ?

= 4.38 moles of H

Mass of C = 2.19 moles * 12.011 g/mol

                =26.30 g of C

Mass of H = 4.38 moles * 1.0079 g/mol

                  = 4.42 g of H

Total mass of compound = mass of C + mass of H + mass of O

65.76 g = 26.30 g + 4.42 g + mass of O

Mass of O = 65.76 g – 26.30 g – 4.42 g

                  = 35.036 g of O

1. The grams of carbon (C) in 65.76 g of the compound = 26.30 g
2. The grams of hydrogen (H) in 65.76 g of the compound = 4.42 g
3. The grams of oxygen (O) in 65.76 g of the compound =35.036 g

Moles of O = 35.036 g / 15.998 g.mol-1

                   = 2.19 moles of O

So form the above calculation

1. The moles of carbon (C) in 65.76 g of the compound = 2.19 moles

2. The moles of hydrogen (H) in 65.76 g of the compound = 4.38 moles
3. The moles of oxygen (O) in 65.76 g of the compound = 2.19 moles

So moles of C and O both are smallest, so we need to divided each mole by this mole

So , C = 2.19 /2.19 = 1

      H = 4.38 / 2.19 = 2

      O = 2.19 /2.19 = 1

So empirical formula is CH2O

Molecular formula = n * empirical formula

n = molecular formula mass / empirical formula mass

= 90.078 / 30.026

= 3

So molecular formula = 3 * CH2O

                                   = C3H6O3

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