Calculate the mass percent of oxygen in chromium(iii) carbonate. round your answer to the nearest percentage.

Answer is: the mass percent of oxygen is 50,7%.
M(Cr₂(CO₃)₃) = 2 · 52 g/mol + 3 · 12 g/mol + 9 · 16 g/mol.
M(Cr₂(CO₃)₃) = 284 g/mol.
In one mole of chromium(III) catbonate there are nine moles of oxygen.
ω(O) = 9 · 16 g/mol ÷ 284 g/mol.
ω(O) = 0,507 · 100% = 50,7%.

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AsiaWoerner AsiaWoerner · Answer 2 · 2019-06-19T16:49:56+02:00

Answer:

The percentage is 50.7%

Explanation:

The formula of chromium (III) carbonate is (Cr₂(CO₃)₃)

Let us calculate the molar mass of (Cr₂(CO₃)₃)

molar mass of (Cr₂(CO₃)₃) =

2X atomic mass of Cr + 3X atomic mass of C + 9X atomic mass of O

molar mass of (Cr₂(CO₃)₃) = 2X 52 + 3X12 +9X16 = 284g/mol

As per the formula there are nine moles of oxygen in each mole of (Cr₂(CO₃)₃).

so we can say that

In 284g of (Cr₂(CO₃)₃) the mass of oxygen = 144g

so 1 g of (Cr₂(CO₃)₃) the mass of oxygen = [tex]\frac{144}{284}=0.507g[/tex]

Therefore

the mass of oxygen in 100g of(Cr₂(CO₃)₃) =0.507X100=50.7

Thus percentage of Oxygen in (Cr₂(CO₃)₃) =50.7%