Question

A sample was decomposed in the laboratory and found to have 38.67g C, 16.22g H, and 45.11g N.

a)Find the molecular formula of this compound if the Formula mass is 62.12 g/mole.
b)Determine how many H atoms would be in a 3.50g sample of this compound.

Answer 1

To start you convert the masses of each element to moles.
\[38.67 g /12.01 \frac{ g }{ mol C} = 3.22mol C\]
\[16.22g / 1.01 \frac{ g }{ mol H} = 16.06mol H\]
\[45.11g/14.01\frac{ g }{ mol N} = 3.22mol N\]
You then can divide these by the lowest value, the moles of either Carbon or Nitrogen in this case. Round if needed.
\[3.22mol C /3.22mol C = 1.00\]
\[16.06mol H /3.22mol C = 4.99\approx 5.00\]
\[3.22mol N /3.22mol C = 1.00\]
This allows us to build the empirical formula.
\[CNH_5\]
We then compute the molar mass of the empirical formula.
\[1mol C \times 12.01\frac{ g }{ mol C } + 1mol N \times \frac{ g }{ mol N} + 5mol H \times \frac{g}{mol H} = 31.07\]
So we know we need to double the empirical formula to get the molecular weight specified.
\[(CNH_5)_2\]
or
\[C_2N_2H_{10}\]

Answer 2

To determine how many H atoms are in a sample, we need to find out how many moles that sample is, multiply that by the number of H atoms per molecule, and then convert that number from moles to atoms.
\[3.50g / \frac{62.12g}{mol} = 0.0563mol\]
\[0.0563mol \times \frac{10molH}{mol} = 0.563mol\]
\[0.563molH \times 6.02 \times 10^{23} \frac{atoms}{mol} = 3.39 \times 10^{23} atomsH\]