Answer:
To find the percent composition of Sn(OH)2, we need to determine the mass percentage of each element in the compound.
The formula for Sn(OH)2 indicates that it contains one tin atom (Sn), two hydroxide ions (OH), each consisting of one oxygen atom (O) and one hydrogen atom (H). We can use the atomic weights of the elements to calculate the percent composition.
Calculate the molar mass of Sn(OH)2:
Sn: 118.71 g/mol (atomic weight from periodic table)
O: 15.99 g/mol (atomic weight from periodic table)
H: 1.01 g/mol (atomic weight from periodic table)
Molar mass of Sn(OH)2 = (1 x 118.71) + (2 x (15.99 + 1.01)) = 213.72 g/mol
Calculate the mass of each element in 1 mole of Sn(OH)2:
Sn: 118.71 g/mol x (1 mole Sn / 1 mole Sn(OH)2) = 118.71 g Sn
O: 15.99 g/mol x (2 mole O / 1 mole OH) x (1 mole OH / 1 mole Sn(OH)2) x (213.72 g/mol Sn(OH)2) = 24.31 g O
H: 1.01 g/mol x (2 mole H / 1 mole OH) x (1 mole OH / 1 mole Sn(OH)2) x (213.72 g/mol Sn(OH)2) = 4.09 g H
Calculate the percent composition of each element in Sn(OH)2:
% Sn = (mass of Sn / total mass of Sn(OH)2) x 100% = (118.71 g / 213.72 g) x 100% ≈ 55.55%
% O = (mass of O / total mass of Sn(OH)2) x 100% = (24.31 g / 213.72 g) x 100% ≈ 11.38%
% H = (mass of H / total mass of Sn(OH)2) x 100% = (4.09 g / 213.72 g) x 100% ≈ 1.91%
Therefore, the percent composition of Sn(OH)2 is approximately 55.55% tin, 11.38% oxygen, and 1.91% hydrogen.
Step-by-step explanation: