A compound that contains 6.44g of boron and 1.80g of hydrogen has a molar mass of approximately 28g/mol. What is its molecular formula?

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Answer 1 · 2015-10-23T08:06:40

The molecular formula is $B_{2} H_{6}$ $B_2H_6$ .

The empirical formula would be under the form of $B_{x} H_{y}$ $B_xH_y$ . We need to find $x and y$ $x and y$ .

Step 1.
From the masses of $B$ $B$ and $H$ $H$ we will find the number of moles:
$n_B=(6.44cancel(g))/(10.81(cancel(g)/"mol"))=0.596mol$

$n_H=(1.80cancel(g))/(1.01(cancel(g)/"mol"))=1.78mol$

Step 2.
In order to find the molar ratio between $B$ and $H$ we divide both number of moles by the smallest one $0.596 mol$ .

$B: (0.596cancel(mol))/(0.596cancel(mol))=1$

$H: (1.78cancel(mol))/(0.596cancel(mol))=2.99$

Step 3.
Therefore, the empirical formula is $BH_3$ .

Step 4.
The molecular mass is $("empirical formula")_n$ where $n$ is calculated by:
$n=(MM)/("e.f.m.")$

where $MM$ is the molecular formula and $"e.f.m."$ is the empirical formula mass:

$n=(28cancel(g/"mol"))/(13.84cancel(g/"mol"))=2$

Therefore, the molecular formula is $B_2H_6$ .