Question

A 0.1510 gram sample of a hydrocarbon produces 0.5008 gram CO2 and 0.1282 gram H2O in combustion analysis. Its

molecular weight is found to be 106. For this hydrocarbon, determine (a) it‟s percent composition; (b) its empirical
formula; (c) its molecular formula.

Answer 1

To determine the percent composition of the hydrocarbon, we first need to calculate the mass of carbon and hydrogen in the sample. From the combustion analysis, we can obtain the masses of CO2 and H2O produced. By comparing the moles of carbon and hydrogen, we can determine the empirical formula. The molecular formula is found by comparing the empirical formula mass with the given molar mass. Hence the correct answer is option B

Step-by-step explanation:

To determine the percent composition of the hydrocarbon, we first need to calculate the mass of carbon and hydrogen in the sample. From the combustion analysis, we know that 0.5008 grams of CO2 is produced. Since the molar mass of CO2 is 44.01 g/mol, this corresponds to 0.0114 moles of CO2. Similarly, we know that 0.1282 grams of H2O is produced. With the molar mass of H2O being 18.02 g/mol, this corresponds to 0.00713 moles of H2O. From these values, we can calculate the moles of carbon and hydrogen:

Moles of carbon = 0.0114 moles CO2 * 1 mole C / 1 mole CO2 = 0.0114 moles C

Moles of hydrogen = 0.00713 moles H2O * 2 moles H / 1 mole H2O = 0.01426 moles H

Now we divide both values by the smallest number of moles, which is 0.0114 moles:

Moles of carbon = 0.0114 moles C / 0.0114 moles C = 1 mole C

Moles of hydrogen = 0.01426 moles H / 0.0114 moles C = 1.25 moles H

The empirical formula therefore is CH. To find the molecular formula, we need to compare the empirical formula mass (14.03 g/mol) with the given molar mass (106 g/mol). The ratio is 106 g/mol / 14.03 g/mol = 7.56. This means that the molecular formula is 7.56 times the empirical formula, giving us C7.56H7.56. To simplify, we round this to C8H8. Therefore, the molecular formula of the hydrocarbon is C8H8.

Hence the correct answer is option B

Answer 2

In a combustion of a hydrocarbon compound, 2 reactions are happening per element:

C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O

Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.

1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C

Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H

The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(0.1510 g) * 100
Percent composition = 99.88%

2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:

Amount of C = 0.01138 mol
Amount of H = 0.014244 mol

Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25

The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5

Thus, the empirical formula of the hydrocarbon is C₄H₅.

3. The molar mass of the empirical formula is

Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.

Molecular Formula = C₈H₁₀