Question

A gaseous hydrocarbon (a compound that contains only hydrogen and carbon) is found to be 11 % hydrogen by mass. a. Find the empirical formula for the compound.A gaseous hydrocarbon (a compound that contains only hydrogen and carbon) is found to be 11 % hydrogen by mass. a. Find the empirical formula for the compound.

Answer 1

To find the empirical formula of the gaseous hydrocarbon, assume 100 grams of the compound. Since it is 11% hydrogen by mass, it contains 11 grams of hydrogen. The remaining mass is carbon. Divide the moles of each element by the smallest number of moles to get the empirical formula CH₂. The molecular formula cannot be determined without the molar mass.

Step-by-step explanation:

Empirical Formula Calculation

To find the empirical formula, assume 100 grams of the gaseous hydrocarbon. Since it is 11% hydrogen by mass, this means it contains 11 grams of hydrogen. The remaining mass, 89 grams, must be carbon. The molar mass of hydrogen is 1 g/mol, and the molar mass of carbon is 12 g/mol. Divide the moles of each element by the smallest number of moles to get the empirical formula. In this case, 11 g H / 1 g/mol = 11 mol H and 89 g C / 12 g/mol = 7.42 mol C. Dividing both by the smallest number of moles, we get the empirical formula CH₂.

To determine the molecular formula, you'll need the molar mass of the compound. However, this information is not provided in the question. Without the molar mass, it is not possible to determine the molecular formula.

Answer 2

Answer: The empirical formula will be=
`C_2H_3`

Step-by-step explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= (100-11) = 89 g

Mass of H = 11 g

Step 1 : convert given masses into moles.

Moles of C =
`\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (89g)/(12g/mole)=7.4moles`

Moles of H =
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (11g)/(1g/mole)=11moles`

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =
`(7.4)/(7.4)=1`

For H =
`(11)/(7.4)=1.5`

The ratio of C : H: = 1: 1.5

Convertig them into whole number ratios

The ratio of C : H: = 2: 3

Hence the empirical formula is
`C_2H_3`

The empirical formula will be=
`C_2H_3`