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3 years ago

What is the mass percent of oxygen (0) in SO2?

Chemistry

1 answer:

vazorg [7]3 years ago

5 0

Answer:

D. (16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%

Explanation:

Step 1: Detemine the mass of O in SO₂

There are 2 atoms of O in 1 molecule of SO₂. Then,

m(O) = 2 × 16.0 g = 16.0 g + 16.0 g = 32.0 g

Step 2: Determine the mass of SO₂

m(SO₂) = 1 × mS + 2 × mO = 1 × 32.1 g + 2 × 16.0 g = 32.1 g + 16.0 g + 16.0 g = 64.1 g

Step 3: Detemine the mass percent of oxygen in SO₂

We will use the following expression.

m(O)/m(SO₂) × 100%

(16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%

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Answer:

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Explanation:

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4 years ago

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A sample of raw mining ore contains a hydrated salt called copper sulfate tetrahydrate, CuSO4.4H2O, along with other impurities.

julsineya [31]

Answer:

88.5 % of the raw ore is CuSO4*4H2O

Explanation:

Step 1: Data given

Mass of the ore = 10.854 grams

Mass of water = 2.994 grams

Step 2: Calculate moles H2O

Moles H2O = mass H2O / molar mass H2O

Moles H2O = 2.994 grams / 18.02 g/mol

Moles H2O = 0.166 moles

Step 3: Calculate moles CuSO4*4H2O

For 1 mol CuSO4*4H2O we have 4 moles H2O

For 0.166 moles H2O we have 0.166/4 = 0.0415 moles CuSO4*4H2O

Step 4: Calculate mass CuSO4*4H2O

Mass CuSO4*4H2O = moles * molar mass

Mass CusO4*4H2O = 0.0415 moles * 231.67 g/mol

Mass CuSO4*4H2O = 9.61 grams

Step 5: Calculate mass % of CuSO4*4H2O

Mass % = (9.61 grams / 10.854 grams )*100 %

Mass % = 88.5 %

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