PRELIMINARY STAGE 2026
Contest 10
Tema Secondary School: 42 points
St. Paul's SHS, Denu: 31 points
Tarkwa SHS: 29 points
Contest 11
Ghanata SHS: 51 points
O’Reilly SHS: 43 points
Yendi SHS: 22 points
Contest 12
St. Joseph Seminary SHS: 52 points
KNUST SHS: 39 points
SDA SHS, Agona: 28 points
ROUND 2 - SPEED RACE
QUESTION
Determine the empirical formula of an oxide of an element X that contains 30.4% by mass of X, given that the atomic mass of X is 14 g/mol.
NOTE: An atomic mass of 14 identifies X as nitrogen, so the oxide is NO$_2$, not SO$_2$.
The calculation gives NO$_2$, corrected here (likely a mishearing of "N-O-2" as "S-O-2").
ANSWER: NO$_2$
SOLUTION:
$\text{O} = 100-30.4$
$\text{O} = 69.6\%$
$\text{Moles X} = \dfrac{30.4}{14}$
$\text{Moles X} = 2.17$
$\text{moles O} = \dfrac{69.6}{16}$
$\text{moles O} = \dfrac{696\times10^{-1}}{16}$
$\text{moles O} = \dfrac{87}{2} \times10^{-1}$
$\text{moles O} = 4.35$
Ratio O:
$\text{X} = \dfrac{4.35}{2.17}$
$\text{X} \approx 2$
Empirical formula $=$ XO$_2$ $=$ NO$_2$ (since X, atomic mass 14, is nitrogen).
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PRACTICE QUESTIONS
1. QUESTION: Determine the empirical formula of an oxide of an element X that contains 40.0% by mass of X, given that the atomic mass of X is 32 g/mol. (O = 16)
ANSWER: $XO_3$
SOLUTION:
$\%O = 100 - 40.0$
$\%O = 60.0\%$
$\text{moles X} = \dfrac{40.0}{32}$
$\text{moles X} = \dfrac{5}{4}$
$\text{moles X} = 1.25$
$\text{moles O} = \dfrac{60.0}{16}$
$\text{moles O} = \dfrac{15}{4}$
$\text{moles O} = 3.75$
$X : O = 1 : 3$
Empirical formula: $XO_3$
2. QUESTION: Determine the empirical formula of an oxide of an element X that contains 60.0% by mass of X, given that the atomic mass of X is 24 g/mol. (O = 16)
ANSWER: $XO$
SOLUTION:
$\%O = 100 - 60.0$
$\%O = 40.0\%$
$\text{moles X} = \dfrac{60.0}{24}$
$\text{moles X} = \dfrac{5}{2}$
$\text{moles X} = 2.50$
$\text{moles O} = \dfrac{40.0}{16}$
$\text{moles O} = \dfrac{5}{2}$
$\text{moles O} = 2.50$
$X : O = 1 : 1$
Empirical formula: $XO$
3. QUESTION: Determine the empirical formula of an oxide of an element X that contains 60.0% by mass of X, given that the atomic mass of X is 48 g/mol. (O = 16)
ANSWER: $XO_2$
SOLUTION:
$\%O = 100 - 60.0$
$\%O = 40.0\%$
$\text{moles X} = \dfrac{60.0}{48}$
$\text{moles X} = \dfrac{5}{4}$
$\text{moles X} = 1.25$
$\text{moles O} = \dfrac{40.0}{16}$
$\text{moles O} = \dfrac{5}{2}$
$\text{moles O} = 2.50$
$X : O = 1 : 2$
Empirical formula: $XO_2$
4. QUESTION: Determine the empirical formula of an oxide of an element X that contains 40.0% by mass of X, given that the atomic mass of X is 16 g/mol. (O = 16)
ANSWER: $X_2O_3$
SOLUTION:
$\%O = 100 - 40.0$
$\%O = 60.0\%$
$\text{moles X} = \dfrac{40.0}{16}$
$\text{moles X} = \dfrac{5}{2}$
$\text{moles X} = 2.50$
$\text{moles O} = \dfrac{60.0}{16}$
$\text{moles O} = \dfrac{15}{4}$
$\text{moles O} = 3.75$
$X : O = 2 : 3$
Empirical formula: $X_2O_3$
5. QUESTION: Determine the empirical formula of an oxide of an element X that contains 80.0% by mass of X, given that the atomic mass of X is 64 g/mol. (O = 16)
ANSWER: $XO$
SOLUTION:
$\%O = 100 - 80.0$
$\%O = 20.0\%$
$\text{moles X} = \dfrac{80.0}{64}$
$\text{moles X} = \dfrac{5}{4}$
$\text{moles X} = 1.25$
$\text{moles O} = \dfrac{20.0}{16}$
$\text{moles O} = \dfrac{5}{4}$
$\text{moles O} = 1.25$
$X : O = 1 : 1$
Empirical formula: $XO$
6. QUESTION: Determine the empirical formula of an oxide of an element X that contains 60.0% by mass of X, given that the atomic mass of X is 12 g/mol. (O = 16)
ANSWER: $X_2O$
SOLUTION:
$\%O = 100 - 60.0$
$\%O = 40.0\%$
$\text{moles X} = \dfrac{60.0}{12}$
$\text{moles X} = 5.00$
$\text{moles O} = \dfrac{40.0}{16}$
$\text{moles O} = \dfrac{5}{2}$
$\text{moles O} = 2.50$
$X : O = 2 : 1$
Empirical formula: $X_2O$