Question

a sample of 2.416 grams of a compound containing carbon and hydrogen reacts with oxygen at elevated temperatures to yield 7.072 grams of co₂ and 4.344 grams of h₂o. (a) calculate the masses of c and h in the sample. grams: c = g; h = g (b) does the compound contain any other elements? (c) what are the mass percentages of c and h in the compound? mass percentages: c = %; h = % (d) what is the empirical formula of the compound? enter the elements in the order presented in the question. empirical formula =

Explanation:

Step1: Calculate moles of C from CO₂

The molar mass of CO₂ is $M_{CO_2}=12.01 + 2\times16.00=44.01$ g/mol. The mass of C in 7.072 g of CO₂ is calculated as follows:
The number of moles of CO₂, $n_{CO_2}=\frac{m_{CO_2}}{M_{CO_2}}=\frac{7.072}{44.01}$ mol. Since 1 mole of CO₂ contains 1 mole of C, the mass of C, $m_C=\frac{12.01}{44.01}\times7.072 = 1.927$ g.

Step2: Calculate moles of H from H₂O

The molar mass of H₂O is $M_{H_2O}=2\times1.01+16.00 = 18.02$ g/mol. The number of moles of H₂O, $n_{H_2O}=\frac{m_{H_2O}}{M_{H_2O}}=\frac{4.344}{18.02}$ mol. Since 1 mole of H₂O contains 2 moles of H, the mass of H, $m_H=\frac{2\times1.01}{18.02}\times4.344=0.486$ g.

Step3: Check for other elements

The total mass of C and H is $m_{C + H}=1.927+0.486 = 2.413$ g. The mass of the sample is 2.416 g. The small difference (2.416 - 2.413 = 0.003 g) can be due to experimental error. So we assume the compound contains only C and H.

Step4: Calculate mass - percentages

The mass - percentage of C, $\%C=\frac{1.927}{2.416}\times100\%=79.76\%$.
The mass - percentage of H, $\%H=\frac{0.486}{2.416}\times100\% = 20.11\%$.

Step5: Determine the empirical formula

The number of moles of C, $n_C=\frac{1.927}{12.01}=0.161$ mol.
The number of moles of H, $n_H=\frac{0.486}{1.01}=0.481$ mol.
The ratio of moles of H to moles of C is $\frac{n_H}{n_C}=\frac{0.481}{0.161}\approx3$. So the empirical formula is CH₃.

Answer:

(a) C = 1.927 g; H = 0.486 g
(b) No
(c) C = 79.76%; H = 20.11%
(d) CH₃