Question

A rock is 11.0% by mass fayalite (Fe_2SiO_4), 14.4% by mass forsterite (Mg_2SiO_4), and the remainder silicon dioxide. What is the mass percent of each element in the rock? % Fe %Mg %Si %O

Answer 1

Calculate the %composition of each part of the rock.

Let we have a 100-gram rock.

The formula of fayalite is Fe2SiO4. Use the atomic weights of each element to find the % of each component in Fayalite.

Fe₂ = 2 x 55.85 = 111.7 g/mol
Si = 1 x 28.06 = 28.09 g/mol
O₄ = 4 x 16.00 = 64.0 g/mol

molar mass of fayalite (FeSiO4) = 203.8 g/mol

% of Fe in Fe2SiO4 = (111.7 / 203.8) x 100 = 54.82 %

mass of Fe in 100 g of rock = 54.8 % x (11.0 g fayalite / 100 g rock) = 6.028 g Fe

% Si in Fe2SiO4 = (28.09 / 203.8) x 100 = 13.78 %

mass of Si from fayalite = 13.78 x (11.0 g fayalite / 100 g rock) = 1.5158 g Si

% O in fayalite = (64.0 / 203.8) x100 = 31.4 %

mass of Oxygen in rock from fayalite = (11.0 g - g Fe - g Si ) = (11.0 g - 6.028 g - 1.5158 g) = 3.45 g O

ii) Now do the same thing for Mg2SiO4.

Mg2 = 2 x 24.3 = 48.6 g/mol
Si = 28.09 = 28.09 g/mol
O4 = 4 x 16.0 = 64.0 g/mol
molar mass of Mg2SiO4 = 140.7 g/mol

% Mg in Mg2SiO4 = (48.6 / 140.7) x 100 = 34.5 %
mass of Mg in rock = 34.5 x (14.4 g forsterite / 100 g rock) = 4.97 g

% Si in Mg2SiO4 = (28.09 / 140.7) x 100 = 19.96 %
mass of Si in rock from forsterite = 19.96 x (14.4 g forsterite / 100 g rock) = 2.87 g

% of O in MgSiO4 = (64.0 /140.7 ) x 100 = 45.48 %

mass of O in rock = (14.4 g - 4.97 g Mg - 2.87g Si ) = 6.56 g O


The rest of the rock (100 g - 11 g fayalite - 14.4 g forsterite = 74.6 g) is SiO2.

Si = 28.09 = 28.09 g/mol
O2 = 2 x 16.00 = 32.00 g/mol
molar mass of SiO2 = 60.09 g/mol

%Si in SiO2 = (28.09 / 60.09) x 100 = 46.74%
mass of Si in rock = 46.74 x (74.6 g SiO2 / 100 g ) = 34.86 g Si

%O in SiO2 = (100% - 46.76%) = 53.24%
mass of O in rock = 53.24 x (74.6 g SiO2 /100 g) = 39.72 g O

Since the only source of Fe is fayalite, so % Fe = (6.028 g Fe / 100 g) x 100 = 6.028 % Fe

Since the only source of Mg is forsterite, so % Mg = (4.97 g  / 100 g) x 100 = 4.97 % Mg

There is Si in all three sources: total = (34.86 g + 2.87 g + 1.5158 g ) = 39.246 g Si = 39.246 % Si

There is O in all three sources: total = (39.72 g + 6.56 g + 3.45 g) = 49.73 g = 49.73 % O