The empirical formula, <span>C<span>H2</span></span>, has a relative molecular mass of
<span>1×<span>(12.01)</span>+2×<span>(1.01)</span>=14.04</span>
This means that the empirical formula must be multiplied by a factor to bring up its molecular weight to 70. This factor can be calculated as the ratio of the relative masses of the molecular and empirical formulas
<span><span>7014.04</span>=4.98≈5</span>
Remember that subscripts in molecular formulas must be in whole numbers, hence the rounding-off. Finally, the molecular formula is
<span><span>C<span>1×5</span></span><span>H<span>2×5</span></span>=<span>C5</span><span>H<span>10</span></span></span>
If concentration of HCl is 1 mol/dm³ :
m(<span>erlenmeyer flask) = 88,00 g.
m(Zn) = 25,0 g.
V(HCl) = 15 ml = 15 cm</span>³ = 0,015 dm³.
Chemical reaction: Zn + 2HCl → ZnCl₂ + H₂.
n(HCl) = c(HCl) · V(HCl).
n(HCl) = 1 mol/dm³ · 0,015 dm³ = 0,015 dm³.
n(Zn) = 25 g ÷ 65,4 g/mol = 0,38 mol.
n(H₂) = 0,015 mol ÷ 2 = 0,0075 mol.
m(H₂) = 0,0075 mol · 2g/mol = 0,015 g.
<span>The energy needed to remove an electron from an atom is called ionization energy.</span>
The percent composition of the compound.
A O-18.18%, N-21.21%, H-60.60%
<h3>Further explanation</h3>
Given
6.00 grams of oxygen,
7.00 grams of nitrogen,
20.00 grams of hydrogen.
Required
The percent composition
Solution
Total mass :
= mass of O + mass of N + mass of H
= 6 + 7 + 20
= 33 g
% O = 6/33 x 100%= 18.18%
% N = 7/33 x 100%=21.21%
% H = 20/33 x 100% = 60.6 %
