Answer is: <span>c. Fe</span>₃<span>O</span>₄<span>.
</span>ω(Fe) = 72,360%.
ω(O) = 100% - 72,36% = 27,64%.
For example, if we the mass of compound is 100 g:
m(Fe) = 72,36 g.
n(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 72,36 g ÷ 55,85 g/mol.
n(Fe) = 1,296 mol.
n(O) = 27,64 g ÷ 16 g/mol.
n(O) = 1,727 mol.
n(Fe) : n(O) = 1,296 mol : 1,727 mol.
n(Fe) : n(O) = 1 : 1,33 or 3 : 4.
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Parents SHOULDN'T avoid sharing their own mistakes with children.
Answer:
23.34 %.
Explanation:
- The percentage of water must be calculated as a mass percent.
- We need to find the mass of water, and the total mass in one mole of the compound. For that we need to use the atomic masses of each element and take in consideration the number of atoms of each element in the formula unit.
- <em>Atomic masses of the elements:</em>
Cd: 112.411 g/mol, N: 14.0067 g/mol, O: 15.999 g/mol, and H: 1.008 g/mol.
- <em>Mass of the formula unit:</em>
Cd(NO₃)₂•4H₂O
mass of the formula unit = (At. mass of Cd) + 2(At. mass of N) + 10(At. mass of O) + 8(At. mass of H) = (112.411 g/mol) + 2(14.0067 g/mol) + 10(15.999 g/mol) + 8(1.008 g/mol) = 308.5 g/mol.
- <em> Mass of water in the formula unit:</em>
<em>mass of water</em> = (4 × 2 × 1.008 g/mol) + (4 × 15.999 g/mol) = 72.0 g/mol.
- <em>So, the percent of water in the compound = [mass of water / mass of the formula unit] × 100 = [(72.0 g/mol)/(308.5 g/mol)] × 100 = 23.34 %</em>
