Step one calculate the moles of each element
that is moles= % composition/molar mass
molar mass of Ca = 40g/mol, S= 32 g/mol , O= 16 g/mol
moles of Ca = 29.4 /40g/mol=0.735 moles, S= 23.5/32 =0.734 moles, O= 47.1/16= 2.94 moles
calculate the mole ratio by dividing each mole with smallest mole that is 0.734
Ca= 0.735/0.734= 1, S= 0.734/0.734 =1, O = 2.94/ 0.734= 4
therefore the emipical formula = CaSO4
Answer:
molar mass of methane CH4
= C + 4 H
= 12.0 + 4 x 1.008
= 12.0 + 4.032
= 16.042g/mol
7.31 x 10^25 molecules x 1 mole CH4 = 121.43 moles
6.02 x 10^23 CH4 molecules
121.43 moles CH4 are present.
Explanation:
not to certain if this is right or not.. but hope it helps!
