A 45.8 mg sample of silicon reacts with hydrogen to form 50.7 mg of the compound. What is the empirical formula of the silicon hydride?

Mar 6, 2016

The empirical formula is ${\text{SiH}}_{3}$.

Explanation:

color(white)(mll)"Si"color(white)(ll) +color(white)(ll) "H"_2 color(white)(l)→color(white)(l) ("Si, H")
$\text{45.8 mg"color(white)(m) xcolor(white)(l) "mg"color(white)(mm) "50.7 mg}$

$\text{Mass of H" = xcolor(white)(l) "mg" = "mass of (Si, H) – mass of Si}$

$\text{= 50.7 mg – 45.8 mg = 4.9 mg}$

Our job is to calculate the ratio of the moles of each element.

$\text{Moles of Si " = 45.8color(red)(cancel(color(black)("mg Si"))) × "1 mmol Si"/(28.08 color(red)(cancel(color(black)("mg Si")))) = "1.631 mmol Si}$

$\text{Moles of H" = 4.9 color(red)(cancel(color(black)("mg H"))) × "1 mmol H"/(1.008 color(red)(cancel(color(black)("mg H")))) = "4.86 mmol H}$

To get the molar ratio, we divide each number of moles by the smallest number
($\text{1.631}$).

From here on, I like to summarize the calculations in a table.

$\text{Element"color(white)(X) "Mass/mg"color(white)(X) "mmol"color(white)(Xll) "Ratio"color(white)(mll)"Integers}$
stackrel(—————————————————-——)(color(white)(m)"Si" color(white)(XXXm)45.8 color(white)(Xmm)"1.631" color(white)(Xm)1color(white)(Xmmmm)1
$\textcolor{w h i t e}{m} \text{H" color(white)(XXXXll)4.9 color(white)(mmm)"4.86} \textcolor{w h i t e}{X m l l} 2.98 \textcolor{w h i t e}{X X X l} 3$

The empirical formula is ${\text{SiH}}_{3}$.