Question

A 1.540 gram sample of an alloy containing only tin and zinc was reacted with excess fluorine gas to produce 2.489 grams in total of a mixture of tin IV fluoride and zinc fluoride. Calculate the percent composition by mass of the two metals in the alloy.

Answer 1

Answer:

Tin: 54.3%

Zinc: 45.7%

Explanation:

The molar masses of the elements are:

Tin: Sn = 117.710 g/mol

Zinc: Zn = 65.409 g/mol

Fluorine: F = 18.998 g/mol

The fluorine gas in excess, so the reaction consumes all the alloy, and all the tin is converted to SnF₄ and all the zinc is converted to ZnF₂. The molar masses of the fluorides are:

SnF₄ = 117.710 + 4*18.998 = 193.702 g/mol

ZnF₂ = 65.409 + 2*18.998 = 103.405 g/mol

If we call x the number of moles of SnF₄, and y the number of moles of ZnF₂, the total mass can be calculated knowing that the mass is the number of moles multiplied by the molar mass:

193.702x + 103.405y = 2.489

The number of moles of Sn is the same as SnF₄ (1:1), and also the number of moles of Zn is the same as ZnF₂ (1:1), so the mass of the alloy:

117.710x + 65.409y = 1.540

if we multiply it by -1.581 and sum with the other equation:

117.710x*(-1.581) + 65.409y*(-1.581) + 193.702x + 103.405y = 1.540*(-1.581) + 2.489

7.60249x = 0.05426

x = 0.0071 mol of Sn

117.710*0.0071 + 65.409y = 1.540

65.409y = 0.704259

y = 0.0108 mol of Zn

The masses are the molar mass multiplied by the number of moles:

Sn: 117.710*0.0071 = 0.836 g

Zn: 65.409*0.0108 = 0.704 g

The percent composition is the mass of the substance divided by the total mass multiplied by 100%:

Sn: (0.836/1.540)*100% = 54.3%

Zn: (0.704/1.540)*100% = 45.7%