Question

Caffeine, a molecule found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 10.0 g of caffeine produces 18.13 g of CO₂, 4.639 g of H₂O, and 2.885 g of N₂. Determine the molar mass of the compound if it is between 150 and 210 g/mol.

Answer 1

Answer:

194 g/mol.

Explanation:

Hello,

In this case, one first must compute the mass of each element as shown below:

$C=18.13gCO_2*\frac{12gC}{44gCO_2} =4.945gC\\H=4.639gH_2O*\frac{2.016gH}{18.0152gH_2O}=0.519gH\\N=2.885gN_2\\O=10.0g-4.945g-0.519g-2.885g=1.651gO$

Next, the corresponding moles:

$C=4.945gC*\frac{1molC}{12gC}=0.412mol\\H=0.519gH*\frac{1molH}{1gH}=0.519mol\\N=2.885gN*\frac{1molN}{14gN}=0.206molN\\O=1.648gO*\frac{1molO}{16gO} =0.103molO$

Then, each element's subscripts is found to be:

$C=\frac{0.412}{0.103}=4\\H=\frac{0.519}{0.103}=5\\N=\frac{0.206}{0.103} =2\\O=\frac{0.103}{0.103}=1$

Therefore, the empirical formula is:

$C_4H_5N_2O$

Nonetheless, it has a molar mass of 97bg/mol, thereby, by multiplying such formula by 2 one gets:

$C_8H_10N_4O_2$

Which has a molar mass of 194 g/mol being correctly contained in the given interval.

Best regards.

Answer 2

The molar mass of the compound is found by finding the empirical and

molecular formula of the compound.

• The molar mass of the compound, MM = 2 × 97 g/mol = 194 g/mol

Reasons:

Molar mass of CO₂ = 44.01 g/mol

Number of moles of CO₂ produced = $\dfrac{18.13}{44.01}$ ≈ 0.412 moles

Number of moles of produced C = 0.412 moles

Mass of C = 12 × 0.412 = 4.944 g

Molar mass of H₂O = 18.015 g/mol

Moles of H₂O produced = $\dfrac{4.639}{18.015 }$ = 0.2575 moles

• Moles of produced H = 2 × 0.2575 = 0.515 moles

• Mass of H = 1.00784 × 0.515 ≈ 0.519 g

Molar mass of N₂ = 28.0134 g/mol

• Moles of N produced = 2 × $\dfrac{2.885}{28.0134 }$ = 2 × 0.103 = 0.206 moles

Mass of oxygen = 10 - 4.944 - 2.885 - 0.519 = 1.652

• Moles of oxygen, O = $\dfrac{1.652 \, g}{16 \, g/mol}$ ≈ 0.103 moles

Therefore, we get;

Number of moles of produced C = 0.412 moles

Number of moles of produced H = 0.515 moles

Number of moles of oxygen, O ≈ 0.103 moles

Number of moles of N produced = 0.206 moles

Dividing by 0.103 gives;

• Mole ratio of C = $\dfrac{0.412}{0.103} = 4$

• Mole ratio of H = $\dfrac{0.515}{0.103} = 5$

• Mole ratio of O = 1

• Mole ratio of N = $\dfrac{0.206}{0.103} = 2$

• The empirical formula of the compound is therefore; C₄H₅N₂O

• The general molecular formula is of the form (C₄H₅N₂O)ₙ

Molar mass of the compound is between 150 g/mol and 210 g/mol (given)

The molar mass of C₄H₅N₂O = 4×12 + 5×1.00784 + 2×14 + 16 ≈ 97

The molar mass of C₄H₅N₂O ≈ 97 g/mol

Molar mass of the compound is between 150 and 210 g/mol, therefore, n in

(C₄H₅N₂O)ₙ = 2, which gives;

• The molar mass of the compound, MM = 2 × 97 g/mol = 194 g/mol

Learn more here:

https://brainly.com/question/12238420