Find Mass

Find mass #1

Wes Prosser 15 Sep 2011 03:08

Q: If a procedure can deposit 0.115 mol of pure Ca₃(PO₄)₂ on an implant, what is the mass of the coating? Molar mass of Ca is 40.08 g mol^-1, of P is 30.97 g mol^-1 and of O is 16.00 g mol^-1.
(Worksheet 1)

M(Ca₃(PO₄)₂) = 3 × 40.08 + 2 × 30.97 + 8 × 16.00 = 310.2 g mol^-1
m = M × n = 310.2 × 0.115 = 35.7 g

✓ All good!

Last edited on 15 Sep 2011 03:20 by Wes Prosser + Show more

Find mass #2

Wes Prosser 15 Sep 2011 03:13

Q: Determine the mass of 0.125 mol of Na₂CO₃? M(Na) = 22.98 g mol^-1, M(C) = 12.01 g mol^-1, M(O) = 16.00 g mol^-1.
(Worksheet 2)

M(Na₂CO₃) = 2 × 22.98 + 1 × 12.01 + 3 × 16.00 = 105.97 g mol^-1
m = M × n = 105.97 × 0.125 = 13.25 g

✓ All good!

Last edited on 15 Sep 2011 03:20 by Wes Prosser + Show more

Find mass #3

Wes Prosser 15 Sep 2011 03:24

Q: Chlorophyll has a formula C₅₅H₂₂MgN₄O₅. A sample has 0.0011 g of Mg. What mass of C was present in the sample? (M(C) = 12.01 g mol^-1, M(Mg) = 24.31 g mol^-1).
(Worksheet 3)

n = m ÷ M = 0.0011 ÷ 24.31 = 4.5×10^-5
Mole ratio of C:Mg = 55:1
n(C) = 2.5×10^-3
m(C) = M × n = 12.01 × 2.5×10^-3 = 0.030 g

✓ All good!

Last edited on 15 Sep 2011 03:25 by Wes Prosser + Show more

Find mass #4

Wes Prosser 15 Sep 2011 03:30

Q: What mass of iron is required to react with 25.6 g of O to make Fe₂O₃?
(Worksheet 3)

Ratio of Fe:O = 2:3
m(Fe) = 2/3 × m(O) = 2/3 × 25.6 g = 17.07 g

✗ It looks like you were on the right track first up, but crossed that out. The ratio 2:3 is a molar ratio, not a mass ratio. However, your answer would have been correct if an atom of iron weighed the same as an atom of oxygen.

n(O) = m(O) ÷ M(O) = 25.6 g ÷ 16.00 g mol^-1 = 1.6 mol
n(Fe) = n(O) × 2 ÷ 3 = 1.067 mol
m(Fe) = n(Fe) × M(Fe) = 55.85 × 1.067 = 59.57 g

Last edited on 15 Sep 2011 03:31 by Wes Prosser + Show more

Find mass #5

Wes Prosser 15 Sep 2011 03:33

Q: What mass of Fe is in a 15 g sample of Fe₂O₃?
(Worksheet 4)

Because mole ratio = 2:3, must be 6g as 6 + 9 = 15 and is in a ratio of 2:3.

✗ Like the previous question, the 2:3 ratio is a mole ratio, not a mass ratio.

M(Fe₂O₃) = 2 × 55.85 + 3 × 16.00 = 159.7 g mol^-1
n(Fe₂O₃) = m ÷ M = 15 ÷ 159.7 = 0.09393 mol
n(Fe) = 2 × n(Fe₂O₃) = 2 × 0.09393 mol = 0.1879 mol
m(Fe) = n(Fe) × M(Fe) = 0.1879 × 55.85 = 10.5 g

Last edited on 15 Sep 2011 03:34 by Wes Prosser + Show more

Stoichiometry questions in general

Wes Prosser 15 Sep 2011 04:30

Generally speaking, the thing that links the different chemicals together in a chem question is the reaction formula. Because this is always written in molar ratios, the only way to link quantities for one chemical to another is through the number of moles. I imagine this graphically like this:

Example formula: Mg _(s) + 2HCl _(aq) → Mg²⁺ _(aq) + 2Cl^- _(aq) + H_{2 (g)}

species:	Mg		HCl		Mg2+		Cl-		H2
mass:	m(Mg)		m(HCl)		m(Mg2+)		m(Cl-)		m(H2)
	⇅		⇅		⇅		⇅		⇅
moles:	n(Mg)	⇄	n(HCl)	⇄	n(Mg2+)	⇄	n(Cl-)	⇄	n(H2)
	⇅		⇅		⇅		⇅		⇅
concentration:	c(Mg)		c(HCl)		c(Mg2+)		c(Cl-)		c(H2)
	⇅		⇅		⇅		⇅		⇅
volume:	v(Mg)		v(HCl)		v(Mg2+)		v(Cl-)		v(H2)

You can go up and down as much as you like (assuming the quantity exists/makes sense), but you can only go sideways using moles. Using this idea, you can "track a path" of calculations necessary to get from what you've got to what you're trying to find.

Last edited on 15 Sep 2011 04:39 by Wes Prosser + Show more