Question

a compound contains 28% h and 72% c by mass. which of the following is the correct empirical formula for the compound?
a $\ce{c_{3}h_{14}}$
b $\ce{ch}$
c $\ce{c_{2}h_{7}}$
d $\ce{c_{2}h_{5}}$

Explanation:

Step1: Assume mass of compound is 100g

So, mass of C = 72g, mass of H = 28g.

Step2: Calculate moles of C and H

Moles of C: $n_C = \frac{72g}{12g/mol} = 6mol$
Moles of H: $n_H = \frac{28g}{1g/mol} = 28mol$

Step3: Find mole ratio (divide by smallest mole)

Ratio of C:H = $\frac{6}{6} : \frac{28}{6} \approx 1:4.67$ (incorrect, re - check: Wait, maybe I made a mistake. Wait, no, let's recalculate. Wait, 72% C and 28% H. Wait, 72g C: molar mass of C is 12g/mol, so 72/12 = 6. 28g H: 28/1 = 28. Now, divide by the smallest, which is 6? No, wait, maybe the percentages are wrong? Wait, no, the problem says 28% H and 72% C. Wait, maybe I misread. Wait, 28% H and 72% C. So mass of C is 72g, H is 28g. Moles of C: 72/12 = 6, moles of H:28/1 = 28. Now, find the simplest ratio. Let's divide both by 2: 3 and 14. Wait, option a is $C_3H_{14}$? Wait, no, that can't be. Wait, maybe the percentages are 7.2% C and 2.8% H? No, the problem says 28% H and 72% C. Wait, maybe I made a mistake in calculation. Wait, let's check the options. Option c is $C_2H_7$, option d is $C_2H_5$, option b is CH, option a is $C_3H_{14}$. Wait, let's recalculate the mole ratio correctly. Wait, 72g C: 72/12 = 6 moles. 28g H:28/1 = 28 moles. Now, divide 6 and 28 by 2: 3 and 14. So the formula would be $C_3H_{14}$? But that seems odd. Wait, maybe the problem has a typo, but according to the calculation, the mole ratio of C to H is 3:14, so the empirical formula is $C_3H_{14}$, which is option a.

Answer:

a. $C_3H_{14}$