A 1.345-g sample of a compound of barium and oxygen was dissolved in hydrochloric acid to give a solution of barium ion, which was then precipitated with an excess of potassium chromate to give 2.012 g of barium chromate, BaCrO4. What is the formula of the compound

The empirical formula is BaO2

Explanation:

Step 1: Data given

Mass of compound = 1.345 grams

Mass of barium chromate produced = 2.012 grams

Molar mass BaCrO4 = 253.37 g/mol

Step 2: Calculate moles BaCrO4

Moles BaCrO4 = mass BaCrO4 / molar mass BaCrO4

Moles BaCrO4 = 2.012 grams / 253.37 g/mol

Moles BaCrO4 = 0.00794 moles

Step 3: Calculate moles Ba

In 1 mol BaCrO4 we have 1 mol Ba

In 0.00794 moles BaCrO4 we have 0.00794 moles Ba

Step 4: Calculate mass Ba

Mass Ba = moles Ba * molar mass Ba

Mass Ba = 0.00794 moles * 137.33 g/mol

Mass Ba = 1.090 grams

Step 5: Calculate mass O

Mass O = mass compound - mass Ba

Mass O = 1.345 grams - 1.090 grams

Mass O = 0.255 grams

Step 6: Calculate moles O

Moles O = 0.255 grams / 16.0 g/mol

Moles O = 0.0159 moles

Step 7: Calculate mol ratio

We divide by the smallest amount of moles

Ba: 0.00794/ 0.00794 = 1

O: 0.0159/ 0.00794 = 2

This means for 1 Ba atom we have 2 O atoms

The empirical formula is BaO2

b) mutualism

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