Number of O atoms in 7.3 x10^-3 g of CaSO4 x 2H2O

Question

Adrien Duran

Chemistry

6 February, 01:12

Number of O atoms in 7.3 x10^-3 g of CaSO4 x 2H2O

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Answers (1)

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Deandre Travis · Answer 1

The number of O atoms in 7.3 x10 ^-3 g of CaSO4 x 2H2O are = 1.535 x10^20 atoms

calculation

step 1: calculate the moles of CaSO4. 2H2O

moles = mass/molar mass molar mass is of CaSO4. H2O=Mw of Ca + S + O + H2O = 40+32 + (4x16) + 2 ((1x2) + 16) = 172 g/mol

moles is therefore = 7.3 x10^-3 g/172 g/mol = 4.244 x10^-5 moles find the moles of O since there are 6 atoms of O the moles of O = 6 x (4.244 x10^-5) = 0.000255 moles

by use of Avogadro's law constant find the number of O atom that is 1 mole = 6.02 x10 ^23 atoms

0.000255 moles=? atoms

by cross multiplication = 0.000255 moles x (6.02 x10^23 atoms) / 1 mole = 1.535 x10^20 atoms