A mixture was found to contain 1.05g of SiO2, 0.69g of cellulose, 1.82g of calcium carbonate. The percentage of calcium carbonate is in the is [x]%. Note: round to the nearest percent.

The porcentage of calcium carbonate in the mixture is calculated as the weight of calcium carbonate divided by the total wieght of the mixture times 100

% calcium carbonate = 1.82 g * 100 / [ 1.05 g + 0.69 g + 1.82 g] = 182 / 3.56 = 51.12 %

Mass mixture = mass SiO2 + mass cellulose + mass CaCO3 =

mass mixture = 1.05 g + 0.69g + 1.82g = 3.56g

mass CaCO3 = 1.82 g

%CaCO3 = mass CaCO3/mass mixture * 100 = 1.82/3.56 * 100 = 51.123% = 51%