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5. A 0.565-8 sample of cobalt metal reacted with excess sulfur powder to give 1.027 g of cobalt sulfide

5. A 0.565-8 sample of cobalt metal reacted with excess sulfur powder to give 1.027 g of cobalt sulfide Calculate the empirical formula of the product. 

6. A 0.750-g sample of tin foil was heated in air and reacted with oxygen gas to give 0.953 g of tin oxide. Calculate the empirical formula of the product. 

7. (optional) A 1.000-g sample of red phosphorus powder was burned in air and reacted with oxygen gas to give 2.291 g of phosphorus oxide. Calculate the empirical formula and molecular formula of phosphorus oxide given the molar mass is approximately 282 g/mol? 

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5)

we have mass of each elements as:

Co: 0.565 g

S: 1.027 g - 0.565 g = 0.462 g

Divide by molar mass to get number of moles of each:

Co: 0.565/58.93 = 9.588*10^-3

S: 0.462/32.07 = 1.441*10^-2

Divide by smallest:

Co: 9.588*10^-3/9.588*10^-3 = 1

S: 1.441*10^-2/9.588*10^-3 = 1.5

Multiply by 2 to get simplest whole number ratio:

Co: 1*2 = 2

S: 1.5*2 = 3

So empirical formula is:Co2S3

Answer: Co2S3

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