Answer:
29.41% of Calcium and 47.04% of Oxygen
Explanation:
The percent composition of an atom in a molecule is defined as 100 times the ratio between the mass of the atom and the mass of the molecule.
The mass of the molecule of the problem (Ore) is 46.28g. That means the percent composition of Calcium is:
13.61g / 46.28g * 100 = 29.41% of Calcium
And percent composition of Oxygen is:
21.77g / 46.28g * 100 = 47.04% of Oxygen
Answer:
Explanation:
C = 49.48
H = 5.19
O = 16.48
N = 28.85
ratio of moles
= 49.48 / 12 : 5.19 / 1 : 16.48 / 16 : 28.85 / 14
= 4.123 : 5.19 : 1.03 : 2.06
= 4 : 5 : 1 : 2
so the empirical formula = C₄ H₅O N₂
Let molecular formula = ( C₄ H₅ON₂ )ₙ ,
n ( 48 + 5 + 16 + 28 ) = 119.19
97 n = 194.19
n = 2 ( approx )
molecular formula = C₈ H₁₀O₂ N₄
Answer:
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