YEAR 12 HOMEWORK ANSWERS

SHEET 4 - Mole Theory, Precipitation & GRAVIMETRIC ANALYSIS

 

Q1. How many significant figures are in the following

(a) 123

3

(b) 1230

3

(c) 1.230

4

(d) 1000

1

(e) 0.0001

1

(f) 0.0100

3

(g) 1.2 x102

2

(h) 106

1

(i)1.2 x 102-

2

(j) 6.02 x 1023

3

Q2. Calculate the RAM of element "X" if it has 3 natural isotopes: the first has a RIM of 31.97 and an abundance of 95.1%, the second has a RIM of 32.97 and abundance of 0.70% and the last has a RIM of 33.97 and an abundance of 4.2%. What is the element?

 

RAM = sum of (RIM x abundances) / total abundance

= (31.97 x 95.1 + 32.97 x .70 + 33.97 x 4.2)/100

= 32.061

Looking up the periodic table - Sulfur has a RAM of 32.06, thus the element is sulfur

Q3. Calculate the molar mass and percentage compositions of the following compounds:

    1. H2O M(H2O) = 1 x 16 + 2 x 1.0 = 18

      2. %H = 2 x 1/18 x 100/1% = 11.1%

      %O = 16/18 x 100/1 = 88.9%

    2. CuSO4 M(CuSO4) = 63.5 + 32.1 + 4 x 16 = 159.6

      3. %Cu = 63.5/159.6 x 100/1% = 39.8%

      %S = 32.1/159.6 x 100/1% = 20.1%

      %O = 4 x 16/159.6 x 100/1% = 40.1%

    3. H2SO4 M(H2SO4) = 2 x 1 + 32.1 + 4 x 16 = 98.1

      4. %H = 2 x 1/98.1 x 100/1% = 2%

      %S = 32.1/98.1 x 100/1% = 32.7%

      %O = 4 x 16/98.1 x 100/1% = 65.2%

    4. C6H12O6 M(C6H12O6) = 6 x 12 + 12 X 1 + 6 x 16 = 180

%C = 6 x 12/180 x 100/1% = 40%

%H = 12 x 1/180 x 100/1% = 6.7%

%O = 6 x 16/180 x 100/1% = 53.3%

Q4. A 0.8361g sample of an organic compound is analysed and found to contain 0.4565g of carbon and 0.0761g of Hydrogen, with the rest of the mass being made up of oxygen. The molar mass of the compound is 88 g/mol. What is the molecular formula of the compound?

 

mOxygen = 0.8361 - (0.4565 + 0.0761) = 0.3035g

 
 

C:H:O

Mass Ratio =

0.4565: 0.0761: 0.3035

Mole ratio =

0.4565/12: 0.0761/1: 0.3035/16

=

0.0380: 0.0761: 0.0190

Lowest ratio =

2:4:1

 So the empirical formula is C2H4O1

M (C2H4O1) = 2 x 12 + 4 x 1 + 1 x 16 = 44 g/mol

Thus we must multiply the M by 2 to get the molecular formula which must be C4H8O2

 

Q5. Calculate the amount of substance of:

  1. 30.0g of water

    2. Using n=m/M

    n = 30.0/(2 x 1.0 + 16.0)

    = 1.67 mol (3 sf)

  2. 51.0g of ammonia

    3. Using n = m/M

    n = 51.0 / 3 x 1.0 + 14.0

    = 3.00 (3 sf)

  3. 5kg of iron (III) oxide

    4. Iron (III) oxide is Fe2O3

    Using n = m/M

    n = 5000/ (2 x 55.8 + 3 x 16)

    = 31.3

    = 30 mol (1 sf)

  4. 25.0 ml of 0.05 M HCl solution

    5. Using n = cV

    n = 0.05 x 25.0/1000 (or 0.025)

    = 0.0012 mol

  5. 3.6L of 0.56M NaNO3 solution

    6. Using n = cV

    n = 0.56 x 3.6

    = 2.0 mol (2 sf)

  6. 1.2 x 1022 molecules of methane

Using n = No particles/NA

n = 1.2 x 1022 / 6.0 x 1023

= 0.020 (2 sf)

(g) 2.0 L of carbon dioxide gas at STP

Using n = VSTP/ 22.4

n = 2.0 / 22.4

= 0.089 mol

(h) 300 ml of hydrogen gas at 40 oC and 90 kPa

Using pV = nRT where V = 0.3 L (ie we use the 90 kPa) and T = 313K

n = pV/RT

= 90 x 0.3/(8.31 x 313)

= 0.01 mol (I SF)

 

Q6. Calculate the mass of

(a) 1.26 mol of V2O5

Using n = m/M so m = n x M

m = 1.25 x (2 x 50.9 + 5 x 16)

= 1.25 x 181.8

= 227g
  1. 0.05 mol of K2Cr2O7

    2. Using m = n x M

    M = 0.05 x (2 x 39.1 + 2 x 52 + 7 x 16)

    = 0.05 x 294.2

    = 15 g (OK it really should be 1 sf)

  2. 1.65 mol of ethene

    3. Ethene is C2H4

    Using m = n x M

    m = 1.65 x (2 x 12 + 4 x 1)

    = 1.65 x 28

    = 46.2 g

  3. Al(NO3)3 in 50.0 ml of 0.50 mol solution

    4. You first need to find n from n = cV, then substitute this into m = nxM

    n = 0.50 x 50.0/1000

    = 0.025 mol

    m = n x M

    = 0.025 x (107.9 + 3 x 14.0 + 9 x 16.0)

    = 0.025 x 293.9

    = 7.3 g

  4. 1.0 x 10-3 mol of Platinum

Using m = n x M

m = 1.0 x 10-3 x 195.1

= 0.20 g (2 sf)

(f ) 4.5 x 1024 molecules of SO2

You first need to find n using n = No particles/ NA then use m = nM

nSO2 = 4.5 x 1024 / 6.0 x 1023

= 7.5 mol

m = n x M

= 7.5 x (32.1 + 2 x 16)

= 7.5 x 64.1

= 480 g
  1. 3 L of Neon gas at 0 oC and 760 mmHg

    2. You first need to find n using n = VSTP / 22.4, then use m=nM

    n = 3/22.4

    = 0.134 mol

    m = n x M

    = 0.134 x 20.2

    = 2.7 g (should be 3 g)

  2. 25 ml of Xenon at 1.4 atm and 30 oC

You first need to find n using pV=nRT, then use m=nM

1.4 atm = 141.9 kPa

30 oC = 303K

25 ml = 0.025 L

n = 141.9 x 0.025/(8.31 x 303)

= 0.00141 mol

m = n x M

= 0.00141 x 131.3

= 0.18 g

 

Q7 Write the chemical formula for the following compounds:

(a) Sodium carbonate

Na2CO3

(b) Iron (II) iodide

FeI2

(c) Barium Sulphate

BaSO4

(d) Aluminium Sulphate

Al2SO4

(e) Tin (II) chloride

SnCl2

(f) Chromium (III) fluoride

CrF3

(g) Calcium oxide

CaO

(h) Silver Nitrate

AgNO3

(i) Carbon dioxide

CO2

(j) Magnesium chloride

MgCl2

(k) Copper (II) acetate

Cu(CH3COO)2

(l) ethanol

CH3COOH

(m) Sodium dichromate

Na2Cr2O7

(n) Calcium hydroxide

Ca(OH)2

(o) potassium nitrate

KNO3

(q) Lead (II) chloride

PbCl2

(r) Iron(III) carbonate

Fe2(CO3)3

(s) Aluminium oxide

Al2O3

(t) Potassium hydrogen carbonate

KHCO3

(u) Silver nitride

Ag3NO3

(v) Calcium phosphate

Ca3(PO4)3

(w) Lithium oxide

Li2O

(x) copper (II) sulphate

CuSO4

(y) copper (II) hydrogen sulphate

H2SO4

(z) Zinc dihydrogen phosphate.

Zn(H2PO4)2

    

 

Q8. Which of the compounds in Q7 are insoluble in water?

(c) Barium Sulphate BaSO4

(n) Calcium hydroxide Ca(OH)2

(q) Lead (II) chloride PbCl2

(r) Iron(III) carbonate Fe2(CO3)3

(s) Aluminium oxide Al2O3

(v) Calcium phosphate Ca3(PO4)

 

Q9 Write balanced equations (including states) for the following reactions

(a) When an electric current is passed through water, it decomposes into it's constituent elements

2H2O (l)

à

2H2(g)

+

O2 (g)

(b) A dilute solution of hydrochloric acid dissolves a strip of magnesium to produce hydrogen gas and magnesium chloride

2HCl (aq) + Mg(s)

à MgCl2 (aq) + H2 (g)

 

(c) When solutions of potassium iodide and lead (II) nitrate are mixed, a precipitate forms

2KI (aq) + Pb(NO3)2

à PbI2 (s) + 2KNO3 (aq)
  1. When carbon dioxide is bubbled through calcium hydroxide solution, calcium carbonate precipitates out of solution
    2.

    Ca(OH)2 (aq) + CO2 (g or aq)

    à CaCO3 (s) + H2O (l)
  2. Butane is burned in a plentiful supply of air
    3.

    C4H10 (g) + 13/2 O2 (g)

    à 4 CO2 (g) + 5H2O (g)
  3. Strontium iodide and lithium sulfate solutions are mixed together.

No net reaction

(g) Lead nitrate solution is added to potassium sulfate sulution

Pb(NO3)2 (aq) + K2SO4 (aq)

à PbSO4 (s) + 2KNO3 (aq)

 

Q10. Calculate the mass of magnesium that is burned in excess oxygen when 5.4g of magnesium oxide is produced.

Reaction is Mg + 1/2 O2

à MgO

n(Mg) = m/M = 5.4/24.3 = 0.222 mol

As n(MgO) = n (Mg)

n(MgO) = 0.222 mol

m(MgO) = n(MgO) x M(MgO)

= 0.222 x (24.3 + 16)

= 8.9 g

Q11. Aluminium used to be used in flashbulbs to produce a bright flash of light when it is burned in oxygen to form aluminium oxide.

(a) What mass of aluminium oxide is formed from 6.0 g of powdered aluminium

4Al (s) + 3O2 (g) -à 2Al2O3 (s)

n (Al) = m/M = 6.0/27.0 = 0.222 mol J

from mole ratios

n (Al2O3) = 2/4 n (Al)

= 0.111 mol

m (Al2O3) = n x M

= 0.111 x 102

= 11 g

(b) What volume of oxygen is consumed at 20 oC and 1.5 atm?

From the above equation, the mole ratios are:

n (O2) = 4/3 n (Al)

= 4/3 x 0.222

= 0.296 mol

V = nRT/p

= 0.296 x 8.31 x (273 + 20)/ 1.5 x 101.3

= 4.7L

 

Q12. Silicon carbide is used to coat drill bits to make them wear resistant. It is produced by heating carbon and silicon dioxide in electric ovens. How much silicon carbide would be produced if 1Kg of carbon and 1 Kg of silicon dioxide were reacted according to the equation

3C(s) + SiO2(s) SiC(s) + 2CO2?

One of these reactants will be in excess. Before we proceed, we must find out which one.

n(C ) = 1000/12 = 83.3 mol

from mole ratios - n (SiO2) = 1/3 n (C)

So if C is the limiting reagent, there would be more than 27.8 mol of SiO2 in the oven

In 1 Kg of SiO2 there are 1000/(28.1 + 2 x 16) = 16.6 mol

Thus SiO2 is the limiting reagent and C is in excess. Thus all calculations must be based on the mass of SiO2

n (SiC) = n (SiO2)

= 16.6 mol

m (SiC) = 16.6 x (28.1 + 12)

= 16.6 x 40.1

= 670 g

 

Q13. The sulfur content of coal can be calculated by combusting the coal with solid sodium carbonate and then forming a barium sulfate precipitate. What is the percentage by mass of sulfur in the coal if a 5.00 g sample of coal produces 0.879g of barium sulfate.

We assume that each atom of S in the coal produces a molecule of the sulfate anion, thus:

S à BaSO4

n (BaSO4) = 0.879/(137.3 + 32.1 + 4 x 16)

= 0.879/233.4

= 0.003766

n (S) = n (BaSO4)

= 0.003766

m (S) = 0.003766 x 32.1

= 0.121 g

% (S) in coal = mass coal/ mass coal sample x 100/1% = 0.121/5.00 x 100/1% = 2.42%

 

Q14. Write a flow chart for a possible gravimetric analysis of the salt content in a packet of salt & vinegar potato crisps. Assume that the analysis can be performed by precipitating the chloride ions as silver chloride.

Steps would involve:

    1. Grind sample of chips
    2. Weigh sample on electronic 3 decimal figure balance (eg 5.000 g)
    3. Mix crushed sample in de-ionized water to dissolve the NaCl
    4. Filter the solid waste from the water. Rinse the filter paper to remove all traces of the NaCl. Discard the solid waster and retain the filterate
    5. Precipitate the chloride ions as AgCl by adding an excess of AgNO3
    6. Weigh a dry Gooch filter
    7. Filter the AlCl precipitate. Rinse sparingly with de-ionized water
    8. Dry to constant weight in an oven at 105 oC
    9. Weigh and determine the mass of the precipitate
    10. Perform calculations

 

Q15. The moisture in soil can be determined by heating it at 105 oC until constant mass is obtained. In one experiment the following results were obtained.

Mass of crucible 23.45g

Mass of crucible and soil sample 42.16g

Mass of crucible and dried soil 39.89g

Calculate the percentage of water in the soil.

Mass water = 42.16 - 39.89 = 2.27 g

Mass soil = 42.16 - 23.45 = 18.71g

% water = (mass water/mass soil sample) x 100/1%

= 2.27/18.71 x 100/1 %

= 12.1 %

 

Q16. In order to analyse a silver alloy for it's silver content, 1.25g of the alloy was dissolved in concentrated nitric acid (which converted the silver into silver nitrate). Excess sodium chloride solution was used to precipitate out the silver ions. The mass of the dried precipitate was 1.12g.

(a) Write an equation for the precipitation reaction

Ag+ (aq) + Cl-(aq) à AgCl (s)

(b) Determine the percentage of silver (by mass) in the alloy

n AgCl = 1.12/ (107.9 + 35.5) = 0.00781 mol

n (Ag) = n (AgCl)

= 0.00781 mol

m Ag = n x M

= 0.00781 x 107.9

= 0.843 g

% silver in alloy = (mass silver / mass alloy) x 100/1%

= 0.843 / 1.25 x 100/1%

= 67.4%

(c) Why was an excess of sodium chloride used?

To ensure the complete recovery of any silver ions in solution

 

Q17. In gravimetric analyses, what would be the effect of these errors on the final result?

(a) The precipitate is not dried completely before the final weighing

The mass of the precipitate would be artificially increased, making it seem like there is more of what you are testing for than there really is

(b) The precipitate is dried before it is washed with distilled water

You would cause some of the spectator ions (eg Na+ and NO3-) to crystallize on the precipitate, increasing it's mass (slightly). This would make you think that there is more of what you are testing for than there actually is.

(c) Some of the original sample was spilt after weighing it.

This would artificially decrease the mass of the precipitate, thus making you think there was less of what you were testing for than there actually is present in the sample.

© John Werry March 98