Basic Chemistry Calculations

1. Sep 8, 2008

Maxwell Kraft

Hello. I just started a first year chemistry course after switching to the sciences from the arts, and wanted to double-check some equations with you because this is a little new to me. I'm good with the qualitative stuff, but the quantitative parts come a little harder.

Q1: A sheet of metal is 93.3 mm wide and 40.6 mm long. If it weighs 4.877 g and the density of the metal is 1.74 g/cm3, what is the thickness of the sheet (in mm)?

My answer: 4.877g / 1.74g/cm^3 = 2.80cm^3

93.3mm x 40.6mm = 3780mm (3 sig figures)

2.80cm^3 / 3780mm = 0.740mm

93.3mm x 40.6mm x 0.740mm = 2.80cm^3

Thus, the sheet is 0.740mm thick.

Q2: If PV = [gR(T+273.15)]/M, solve for M when P = 334, V = 0.350, g = 0.274, R = 62.37, and T = 39.

PV = [gR(T+273.15)] / M

(334)(0.350) = (0.274) (62.37) (312.15) / M

116 = 5330 (3 sig figures) / M

116 / 5330 = M

0.0217 = M

Q3: An ore contains 42.3 % of the mineral ilmenite, FeTiO3, which is a source of the element Ti. How much ore must be processed in order to obtain 41.0 kg of Ti?

Molar masses

Fe = 55.85g
Ti = 2004.4g
O = 16.00g x 3 = 48.00g

total = 308.2g

308.2g / 204.4g = 1.50

41.0kg x 1.50 = 61.5kg

100 / 42.3 = 2.36

61.5 x 2.36 = 145 kg of ore are needed to obtain 41.0kg of thallium.

M.

2. Sep 9, 2008

Staff: Mentor

First looks OK. Second looks wrong (don't round down intermediate values, check your math). Third looks wrong (check molar mass of Ti).

3. Sep 9, 2008

Maxwell Kraft

Opps. Got the element wrong for number three. Should be...

Q3: An ore contains 42.3 % of the mineral ilmenite, FeTiO3, which is a source of the element Ti. How much ore must be processed in order to obtain 41.0 kg of Ti?

Molar masses

Fe = 55.85g
Ti = 48.77
O = 16.00g x 3 = 48.00g

total = 151.7g

308.2g / 151.7g = 3.168

41.0kg x 3.168 = 129.9kg

100 / 42.3 = 2.36

129.5 x 2.36 = 300 kg of ore are needed to obtain 41.0kg of titanium.

For number 2, is my mistake that I rounded before I got to a final answer?

4. Sep 9, 2008

Staff: Mentor

One of mistakes. Check your math.

5. Sep 9, 2008

lukas86

6. Sep 10, 2008

Maxwell Kraft

Ok, let me work through this and see if I can figure out where I went wrong.

Q2: If PV = [gR(T+273.15)]/M, solve for M when P = 334, V = 0.350, g = 0.274, R = 62.37, and T = 39.

PV = [gR(T+273.15)] / M

(334)(0.350) = (0.274) (62.37) (312.15) / M

The left side actually works out to 116.9, and I won't round the right side.

116.9 = 5334.5 / M

Moving 5334.5 to the left side, it goes from the numerator to the demoninator.

116.9 / 5334.5 = M

0.0219 = M (3 signifigant figures)

I hope I got that right, because I have another one =P

Q4: A piece of nickel wire has a diameter of 0.505 mm. If nickel has a density of 8.90 g/cc, how long (in meters) should you cut a piece of wire to obtain 0.0247 moles of nickel?

My answer: since a wire is a cyllinder, then the volume = πr2h

r = 0.505 / 2 = 0.252

Nickel weighs 58.69 g/mole, so there are 58.69g/mole x 0.0247 moles = 1.45g of the wire needed to obtain the correct length.

So v = π (0.252)2h

Since densty is 8.90 g/cc, then its volume is 1.45g / 8.90g/cc = 0.163 cc

So 0.163 cc = π (0.252)2h

h = 0.163 cc / π (0.252) 2

h = 0.817cm of wire is needed

I’d like to thank you all for giving me a hand with this. I know this is something you have to practice, but I’m afraid of practicing it wrong and forming bad habits.

7. Sep 10, 2008

Staff: Mentor

Why have you moved M from denominator to numerator? What you are doing you are dividing both sides by 5334.5, it cancels out on the right - but M stays where it was!

Ni wire looks OK to me.

8. Sep 10, 2008

Maxwell Kraft

But, if 116.9 = 5334.5 / M, then doesn't 116.9 / 5334.5 = M? Both sides are divided by 5334.5, so the left side becomes 116.9 / 5334.5 and the right side becomes (5334.5 / M) / 5334.5, which cancels out to just M.

9. Sep 11, 2008

Staff: Mentor

No, it cancels out to 1/M.