Please clearly show steps! Thank you. A. Determine the percent composition of a

ID: 545927 • Letter: P

Question

Please clearly show steps! Thank you.

A. Determine the percent composition of a compound formed from 0.500g of tin reacting with oxygen and forming 0.635 g of the oxide.

B. Determine the empirical and molecular formulas for a compound that has a molar mass of 62g/mol and is 38.7%C, 9.74%H and 51.6%O by mass.

C. Determine the empirical and molecular formula for a compound that is 31.42% S, 31.35% O and 37.23%F and has a molar mass of 102.1g/mol.

D. Determine the number of oxygen atoms in a 45.6 oz. sample of aluminum sulfate.

Element %by mass atomic wt. relative number simple ratio

C 38.7 12 38.7/12 = 3.225 3.225/3.225 = 1.0

H 9.74 1.0 9.74/1.0= 9.74 9.74/3.225 = 3.0

O 51.6 16 51.6/16 =3.225 3.225/3.225 = 1.0

Emperical formula = CH3O

Emperical formula mass = 12+3+16 = 31

Molar mass = 62 grams

n= molar mass/emperical formula mass = 62/31 = 2

Molecular formula= nxemperical formula = 2xCH3O = C2H6O2

Molecular formula = C2H6O2

C).

Element % by mass atomic wt relative number simple ratio

S 31,42 32 31.42/32 = 0.98 0.98/0.98 = 1.0

O 31.35 16 31.35/16 = 1.959 1.959/0.98= 1.99 = 2.0

F 37.23 19 37.23/19 = 1.959 1.959/0.98 = 1.99 = 2.0

Emperical formula= SO2F2

emperical formula mass = 32+32+38=102

Molar mass = 102.1 grams

n= molar mass/emperical formula mass = 102.1/102= 1.0

n= 1.0

Molecular formula = nxemperical formula= 1xSO2F2

Molecular formula = SO2F2.

D).mass of Al2(SO4)3 = 45.6gr ams

molar mass of Al2(SO4)3 =342.15 grams/mole

342.15 grams of Al2(SO4)3 contains = 12x6.023x10^23 atoms of Oxygen

45.6 grams of Al2(SO4)3contains= ?

= 12x6.023x10^23x45.6/342.15 = 9.63x10^23 atoms of oxygen