Question

# a.) How many calcium atoms are in 0.154 g of calcium sulfate? b.) Empirical formula for...

a.) How many calcium atoms are in 0.154 g of calcium sulfate?

b.) Empirical formula for copper chloride?

1) MW of CaSO4 = 136.14 g/gmol

moles CaSO4 = mass/MW = 0.154 g / 136.g/gmol = 0.00113 gmol of CaSO4

n= 0.00113 gmol of CaSO4

1 gmol = 6.022*10^23 molecules (which is avogrado number = A)

molecules in 0.00113 CaSO4

n*A = 0.00113 * 6.022*10^23 = 6.8*10^20

1 molecules of CaSO4 contains 1 atom of Calcium, therefore there are 6.8*10^20 atoms of calcium in 0154 g of that substance

2)Empirical formula for copper chloride

Recall the definition of empirical formula:

empirical formula is the simplest formula for a compound. A molecular formulais the same as or a multiple of the empirical formula, and is based on the actual number of atoms of each type in the compound

Therefore:

CuCl2 is the simplest form! we can't divide since Cu is already 1...

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