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?

1 Answer
Dec 23, 2015

#"45.57% Sn"#

#"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#.

#color(blue)("% Sn" = m_(Sn)/m_"total" xx 100)#

Your initial sample of tine reacts completely with #"24.49 g"# of chlorine, which means that the total mass of the compound will be

#m_"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)#