Question

Determine the molecular formula for a compound with a molar mass of 45 g/mol that contains 79.9% C and 20.1% H.

Answer 1

Determine the molecular formula.

Procedure:

1) You know that this compound's percent composition is as follows

Molar mass = 45 g/mol

C => 79.9%

H => 20.1%

It means:

$\begin{gathered} 45\text{ g Compound }x\text{ }\frac{79.9\text{ g C}}{100\text{ g }Compound}\text{ = 35.95 g C} \\ 45\text{ g Compound x }\frac{20.1\text{ g H}}{100\text{ g Compound}}=9.04\text{ g H} \end{gathered}$

2) Now you have to use the atomic mass in grams of C and H, and determine how many moles of each you need in one mole of your compound.

Atomic masses from the periodic table:

C= 12.01 g/mol

H=1.00 g/mol

For C:

$35.95\text{ g C x }\frac{1\text{ mole C}}{12.01\text{ g C}}\text{ = 2.99 moles = 3 moles (aprrox.)}$

For H:

$9.04\text{ g H x}\frac{1\text{ mole H}}{1.00\text{ g H}}=\text{ 9.04 moles = 9 moles (aprrox.)}$

Now our molecular formula:

$C_3H_9$

(Note: I'm going to skip the empirical formula, I will get directly the molar mass)