Question

Paradichlorobenzene, a common ingredient in solid air fresheners, contains only C, H, and Cl and has a molar mass of about 147 grams. Given that combustion of 1.68 grams of this compound produces 3.02 grams CO2 and 0.412 grams H2O, determine its empirical and molecular formulas.

Answer 1

The empirical formula of the compound is C₂H₂Cl. The molecular formula is approximately C₅H₅Cl₂.

Step-by-step explanation:

To determine the empirical and molecular formulas, we first need to calculate the molar ratios of carbon, hydrogen, and chlorine in the compound. From the given information, we can calculate the mass of each element present in the 1.68 gram sample.

The mass of carbon can be calculated using the molar mass of CO₂ (44 g/mol). Since 1.68 grams of the compound produces 3.02 grams of CO₂, the mass of carbon can be calculated as (3.02 g CO₂) * (12 g C / 44 g CO₂) = 0.822 grams.

Similarly, the mass of hydrogen can be calculated using the molar mass of H₂O (18 g/mol). Since 1.68 grams of the compound produces 0.412 grams of H₂O, the mass of hydrogen can be calculated as (0.412 g H₂O) * (2 g H / 18 g H₂O) = 0.0458 grams.

Lastly, we can calculate the mass of chlorine by subtracting the mass of carbon and hydrogen from the total sample mass: 1.68 g - 0.822 g - 0.0458 g = 0.8122 grams.

Now that we have the mass of each element, we can convert them to moles using their respective molar masses:

moles of carbon = 0.822 g C / 12 g/mol = 0.0685 mol

moles of hydrogen = 0.0458 g H / 1 g/mol = 0.0458 mol

moles of chlorine = 0.8122 g Cl / 35.5 g/mol = 0.0229 mol

Finally, we divide the number of moles of each element by the smallest number of moles to get their relative ratios:

carbon : hydrogen : chlorine = 0.0685 mol / 0.0229 mol : 0.0458 mol / 0.0229 mol : 0.0229 mol / 0.0229 mol = 2 : 2 : 1

Therefore, the empirical formula of the compound is C₂H₂Cl. To determine the molecular formula, we need to know the molecular mass of the compound, which is given as 147 g/mol. The empirical formula mass can be calculated as (2 * 12.01 g/mol) + (2 * 1.01 g/mol) + 35.5 g/mol = 61.56 g/mol. To find the molecular formula, we divide the molecular mass by the empirical formula mass: 147 g/mol / 61.56 g/mol = 2.39 (approximately).

Rounding to the nearest whole number, the molecular formula is approximately C₅H₅Cl₂.

Answer 2

Empirical formula

C3H2Cl

Molecular formula

C6H4Cl2

Step-by-step explanation:

Firstly, we can get the mass of carbon and hydrogen produced by calculating their number of moles from that of carbon iv oxide and water respectively.

To get the number of moles of carbon iv oxide, we simply make a division. The number of moles is the mass of carbon iv oxide divided by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol. The number of moles is thus 3.02/44 = 0.069

One atom of carbon is present in Carbon iv oxide. This means 0.069 moles of carbon is present. The mass of carbon present equals the atomic mass of carbon multiplied by the number of moles = 0.069 * 12 = 0.83g

The mass of water is 0.412g, the number of moles is 0.412/18 = 0.023

In water, there are two atoms of hydrogen. Hence the number of moles of hydrogen is 2 * 0.023 = 0.046

The mass of hydrogen is the number of moles of hydrogen multiplied by the atomic mass unit of hydrogen = 0.046 * 1 = 0.046g

Now we need to know the mass of chlorine. We simply subtract the masses of hydrogen and carbon.

1.68-0.046-0.83 = 0.804

Now we proceed to calculate the empirical formula. We simply divide the mass of each by their atomic masses and then proceed from there.

H = 0.046/1 = 0.046

C = 0.83/12 = 0.069

Cl = 0.804/35.5 = 0.023

We divide by the smallest number which is that of that of carbon

H = 0.046/0.023= 2

C = 0.069/0.023= 3

Cl = 0.023/0.023 = 1

The empirical formula would thus be C3H2Cl

The molecular formula can be calculated using the molar mass;

(C3H2Cl)n = 147

(36+ 2 + 35.5 ) = 147

73.5n = 147

n = 147/73.5 = 2

The molecular formula is thus C6H4Cl2