# Question e0e87

Feb 24, 2016

Here's what I got.

#### Explanation:

The idea here is that you need to use the concept of mass conservation to determine how much silicon and hydrogen were present in your unknown compound.

More specifically, you need to use the fact that the mass of silicon that was present in the compound before the reaction took place will now be present in silicon dioxide, ${\text{SiO}}_{2}$.

Likewise, the mass of hydrogen that was present in the compound before the reaction took place will now be present in water, $\text{H"_2"O}$.

This means that if you can determine how much silicon you get in $\text{31.27 g}$ of silicon dioxide and how much hydrogen you get in $\text{18.72 g}$ of water, you will know the percent composition of silicon and hydrogen in the original compound.

So, in order to find the percent composition of silicon in silicon dioxide, you need to know the molar mass of the compound and the molar mass of silicon.

${\text{For Si: " "28.09 g mol}}^{- 1}$

${\text{For SiO"_2: " 60.08 g mol}}^{- 1}$

Notice that one mole of ${\text{SiO}}_{2}$ contains one mole of silicon, $\text{Si}$, which means that the percent composition of silicon in silicon dioxide will be

(28.09 color(red)(cancel(color(black)("g mol"^(-1)))))/(60.08color(red)(cancel(color(black)("g mol"^(-1))))) xx 100 = "46.75% Si"

This means that every $\text{100 g}$ of silicon dioxide will contain $\text{46.75 g}$ of silicon. Since your sample has a mass of $\text{31.27 g}$, it will contain

31.27color(red)(cancel(color(black)("g SiO"_2))) * "46.75 g Si"/(100color(red)(cancel(color(black)("g SiO"_2)))) = "14.62 g Si"

Now do the same for water. The molar masses of hydrogen and of water are

${\text{For H: " "1.00794 g mol}}^{- 1}$

${\text{For H"_2"O": " 180.15 g mol}}^{- 1}$

Now, notice that one mole of water, $\text{H"_color(red)(2)"O}$, contains $\textcolor{red}{2}$ moles of hydrogen, $\text{H}$. This means that the percent composition of hydrogen in water will be

(color(red)(2) xx 1.00794 color(red)(cancel(color(black)("g mol"^(-1)))))/(18.015color(red)(cancel(color(black)("g mol"^(-1))))) xx 100 = "11.19% H"

So, every $\text{100 g}$ of water will contain $\text{11.19 g}$ of hydrogen, which means that your sample will contain

18.72color(red)(cancel(color(black)("g H"_2"O"))) * "11.19 g H"/(100color(red)(cancel(color(black)("g H"_2"O")))) = "2.095 g H"

So, you know that your original compound contained

• $\text{14.62 g Si}$
• $\text{2.095 g H}$

Use its total mass to find the percent composition of each element

"For Si: " (14.62 color(red)(cancel(color(black)("g"))))/(50color(red)(cancel(color(black)("g")))) xx 100 = color(green)("29% Si")

"For H: " (2.095color(red)(cancel(color(black)("g"))))/(50color(red)(cancel(color(black)("g")))) xx 100 = color(green)("4.2% H")#

I left the answers rounded to two sig figs.