Question

3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 150g/mol is burned completely in excess oxygen, and the mass of the products carefully measured .

Product

Carbon dioxide 5.13g

water 2.10g

Use this information to find the molecular formula of X

Answer 1

C₅H₁₀O₅

Step-by-step explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

`\text{Mass of C} = \text{5.13 g CO}_(2)* \frac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_(2)}= \text{1.400 g C}`

(b) Mass of H

`\text{Mass of H} = \text{2.10 g H$_(2)$O}* \frac{\text{2.016 g H}}{\text{18.02 g H$_(2)$O}} = \text{0.2349 g H}`

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

`\text{Moles of C = 1400  mg C}*\frac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} * \frac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} * \frac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}`

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

`\text{C: } (116.6)/(116.6)= 1\\\\\text{H: } (233.1)/(116.6) = 1.999\\\\\text{O: } (116)/(116.6) = 1.00`

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

`n = \frac{\text{MF Mass}}{\text{EF Mass }} = \frac{\text{150 u}}{\text{30.03 u}} = 5.00  \approx 5`

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.