# If 20.50g of Sn combine completely with 24.49g of Cl to form a compound, what is the percent composition of tin in the compound? What is the empirical formula of the compound?

Dec 23, 2015

$\text{45.57% Sn}$

${\text{SnCl}}_{4}$

#### Explanation:

The percent composition of tin in the compound will be equal to the ratio between the mass of tin and the mass of the compound, multiplied by $100$.

$\textcolor{b l u e}{\text{% Sn" = m_(Sn)/m_"total} \times 100}$

Your initial sample of tine reacts completely with $\text{24.49 g}$ of chlorine, which means that the total mass of the compound will be

${m}_{\text{total" = "20.50 g" + "24.49 g" = "44.99 g}}$

The percent composition of tin in the compound will be

(20.50 color(red)(cancel(color(black)("g"))))/(44.99color(red)(cancel(color(black)("g")))) xx 100 = color(green)("45.57% Sn")

Now, to get the compound's empirical formula, use the molar masses of the two elements to determine how many moles of each you have in that sample

20.50 color(red)(cancel(color(black)("g"))) * "1 mole Sn"/(118.71color(red)(cancel(color(black)("g")))) = "0.17269 moles Sn"

24.49 color(red)(cancel(color(black)("g"))) * "1 mole Cl"/(35.453 color(red)(cancel(color(black)("g")))) = "0.69077 moles Cl"

Divide both values by the smallest one to get the mole ratio that exists between the two elements in the compound

"For Sn: " (0.17269 color(red)(cancel(color(black)("moles"))))/(0.17269 color(red)(cancel(color(black)("moles")))) = 1

"For Cl: " (0.69077 color(red)(cancel(color(black)("moles"))))/(0.17269color(red)(cancel(color(black)("moles")))) ~~ 4

The empirical formula of the compound, which tells you the smallest whole number ratio that exists between the two elements, will be

"Sn"_1"Cl"_4 implies color(green)("SnCl"_4)