Question

Calculate the mass (in grams) of each sample.

3.9×10²⁵ O3 molecules

4.46×10¹⁹ CCl2F2 molecules

9 water molecule(s)

Determine the number of moles of oxygen atoms in each of the following.

2.28×10⁻² mol H2CO3

Give the name from the formula or the formula from the name for each of the following hydrated ionic compounds.

Zn3(PO4)2⋅3H2O

iridium(III) bromide tetrahydrate

BeCrO4⋅5H2O

aluminum nitrate dihydrate

What mass (in grams) of iron(III) oxide contains 67.3 g of iron? Iron(III) oxide is 69.94% iron by mass.

Answer 1

(1) 3.9×10²⁵ O3 molecules

1 mol -------------------> 6.023 x 10^23 molecules

x moles ---------------> 3.9×10²⁵ O3 molecules

x = 3.9×10²⁵ / 6.023 x 10^23

x = 64.75 moles

mole = mass / molar mass

64.75 = mass / 48

mass = 3108 g

(2) 4.46×10¹⁹ CCl2F2 molecules

moles = 4.46×10¹⁹ / 6.023 ×10²³

moles = 7.4 x 10^-5

CF2Cl2 molar mass = 121 g /mol

mass =    121 x 7.4 x 10^-5 = 8.95 x 10^-3 g

(3)   9 water molecule(s)

1mol ----------------------> 6.023 x 10^23 molecules

x ---------------------------> 9 water molecules

x = 9 / 6.023 x 10^23

   = 1.9 x 10^-23

mass = 18 x 1.9 x 10^-23

            = 2.69 x 10^-22 g

(4) Zn3(PO4)2⋅3H2O --------------------> zinc phosphate trihydrate

(5) iridium(III) bromide tetrahydrate ---------------------> IrBr3.4H2O

(6) BeCrO4⋅5H2O ----------------------------> Berillium Chromate pentahydrate

(7) aluminum nitrate dihydrate ----------------> Al(NO3)2 .2H2O

(8 ) What mass (in grams) of iron(III) oxide contains 67.3 g of iron? Iron(III) oxide is 69.94% iron by mass

     Fe2O3

Fe mass % = (Fe mass / total mass ) x 100

69.94 = (67.3 / total mass) x100

total mass = 96.2 g

mass of Fe(III) oxide = 96.2 g