Formula, Total & Net Ionic Equations

To be able to properly write formula, total and net ionic equations, you must know: elements and charges, polyatomic ions and charges, weak and strong acids/bases, solubility rules and be able to balance equations.

FORMULA UNIT EQUATION
This is the type of equation we have been dealing with in previous chapters. Basically it is a displacement reaction.
Step 1: If no products are given, then determine the products.
Step 2: Balance the equation.

Given: lead (II) hydroxide + hydrochloric acid à
1) Pb(OH)2 + HCl à PbCl2 + H2O
2) Pb(OH)2 + 2 HCl à PbCl2 + 2 H2O

TOTAL IONIC EQUATION
As the name states, this equation shows ions (charges) for everything that is soluble or ionizes/dissociates 100%.
Step 3: Look at each compound in the formula unit and ask yourself: Is it soluble? Strong acid? Strong Base?
- If yes to any of the three questions, assign charges to each element in the compound
- If no, then leave the compound as it is

3) Pb(OH)2 + 2H+1 + 2 Cl-1 à Pb+2 + 2 Cl-1 + 2 H2O
(assume HOT water)

Question: Where did the coefficient 2 come from in front of H+1 and Cl-1?
- Coefficient is the number we use to balance equations in the Formula Unit Equation. In the Total Ionic Equation, the coefficients come from 1) the coefficients in the formula unit equation and 2) from the subscript assigned when creating a compound. If you have both a coefficient and a subscript to account for on the same compound, you would multiply the two together:
formula unit equation 2 CaBr2 à total ionic equation 2 Ca+2 + 4 Br-1

Spectator ions do not change their charge (remember ions are charged particles, not compounds). Whatever the charge was on the reactant side it will be the same charge on the product side of the equation. Spectators are present in the solution, but do not participate in the reaction. That is why they have no change in charge. Think of it like a sports fan: if you are a spectator, you are involved in the overall event because you are physically present and viewing the sport, but you do not participate in the action of the sport.

NET IONIC EQUATION
Step 4: Look at each ion (not compound) in the Total Ionic Equation. Cancel out the identical spectator ions that are on both the reactant and product side.
Step 5: What remains (doesn’t cancel) constitutes the Net Ionic Equation. You will always have something that doesn’t cancel. If everything cancels, go back and check your work, something is wrong.
**** Net ionic equations for ALL reactions of strong acids with strong soluble bases that form soluble salts and water is: H+ + OH- à H2O (l)
In other words, if you can’t find anything else, your net ionic equation will be: H+ + OH- à H2O (l)


4) Pb(OH)2 + 2H+1 + 2 Cl-1 à Pb+2 + 2 Cl-1 + 2 H2O
5) Pb(OH)2 + 2H+1 à Pb+2 + 2 H2OPRACTICE PROBLEMS

Write balanced formula unit, total ionic and net ionic equations for the following reactions. Assume all reactions occur in water or in contact with water.

1) acetic acid + sodium hydroxide à



2) calcium hydroxide + hydrosulfuric acid à



3) Ba(NO3)2 + K2CO3 à



4) lead (II) hydroxide + carbonic acid à



5) Our bones are mostly calcium phosphate. Calcium chloride reacts with potassium phosphate to form calcium phosphate and potassium chloride.





ANSWERS

1) Formula Unit: HC2H3O2 + NaOH à NaC2H3O2 + H2O
Total Ionic: HC2H3O2 + Na+1 + OH-1 à Na+1 + C2H3O2-1 + H2O
Weak acid
Net Ionic: HC2H3O2 + OH-1 à C2H3O2-1 + H2O

2) Formula Unit: Ca(OH)2 + H2S à CaS + 2 H2O
Total Ionic: Ca+2 + 2OH-1 + 2H +1 + S-2 à Ca+2 + S-2 + 2 H2O
Net Ionic: 2OH-1 + 2H +1 à 2 H2O
OH-1 + H +1 à H2O

3) Formula Unit: Ba(NO3)2 + K2CO3 à BaCO3 + 2KNO3
Total Ionic: Ba+2 + 2NO3-1 + 2K +1 + CO3-2 à BaCO3 + 2K+1 + 2NO3-1
Insoluble
Net Ionic: Ba+2 + CO3-2 à BaCO3

4) Formula Unit: Pb(OH)2 + H2CO3 à PbCO3 + 2H2O
Total Ionic: Pb(OH)2 + H2CO3 à PbCO3 + 2H2O
Insoluble Weak Acid Insoluble
& Insoluble
Net Ionic: Pb(OH)2 + H2CO3 à PbCO3 + 2H2O

5) Formula Unit: 3CaCl2 + 2K3PO4 à Ca3(PO4)2 + 6KCl
Total Ionic: 3Ca+2 + 6Cl-1 + 6K+1 + 2 PO4-3 à Ca3(PO4)2 + 6K+1 + 6Cl-1
Insoluble
Net Ionic: 3Ca+2 + 2 PO4-3 à Ca3(PO4)2

Formula Writing

Write the correct chemical formula for the following:

1.Magnesium Nitride
2.Iron (III) Oxide
3.Sodium Sulfate
4.Copper (II) Chloride
5.Barium Nitrate
6.Aluminum Hydroxide
7.Mercury (II) Phosphate
8.Aluminum Silicate
9. Copper (II) Bromide
10.Lead (II) Chlorate
11.Silver Cyanide
12.Ammonium Oxide
13.Aluminum Perchlorate
14.Tin (II) Chloride
15.Nickel (III) Acetate
16.Potassium Sulfide
17.Magnesium Bisulfate
18.Iron (II) Phosphate
19.Cobalt (II) Hydrogen Sulfate
20.Chromium (II) Bicarbonate
21.Sodium Hypochlorite
22.Barium Carbonate
23.Zinc (II) Permanganate
ANSWERS
1. Mg3N2
2. Fe2O1
3. Na2SO4
4. CuCl2
5. Ba(NO3)2
6. Al(OH)3
7. Hg3(PO4)2
8. Al2(SiO3)3
9. CuBr2
10. Pb(ClO2)2
11. AgCN
12. (NH4)2O
13. Al(Clo4)3
14. SnCl2
15. Ni(C2H3O2)3
16. K2S
17. Mg(HSO4)2
18. Fe3(PO4)2
19. Co(HSO4)2
20. Cr (HCO3)2
21. NalO
22. BaCo3
23. Zn(MnO4)2

Balancing Equations by the Ion-Electron Method

1. Write the equation to be balanced.
2. Identify what was oxidized and what was reduced.

OXIDATION
REDUCTION
Charge goes up
Charge goes down
Losses electrons
Gains electrons
Gains oxygen
Losses oxygen
Losses hydrogen
Gains hydrogen

3. Write an equation for the oxidized substance and an equation for the reduced substance.
4. Add H+’s, OH-‘s and/or H2O’s to either side of each half-reaction to balance the number of atoms of each type. AFTER, you add H+’s, OH-‘s and/or H2O’s, make sure the equation is balanced.

Atoms to be Balanced
ACIDIC (H+)
BASIC (OH-)
Oxygen
1. Add one H2O for each oxygen, to the side lacking oxygen
2. Add two H+ for each H2O, to the other side for balance
1. Add two OH- for each oxygen (to the side lacking oxygen)
2. Add one H2O to the other side for every two OH-
Hydrogen
1. Add H+
1. Add one H20 to the side needing H+
2. Add one (or more) OH- to the other side for balance.

5. For each half-reaction, add up the total charges on each side and equalize them by adding or subtracting electrons from the left side of the half reaction.
6. Find a least common multiple for the gain or loss of electrons in the two half-reactions and equalize the gain with the loss.
7. Add the two half reactions, canceling the gained and lost electrons (and any extra H2O or H+), and the equations. The equation should be balanced, make a chart to check.

BALANCING EQUATIONS BY INSPECTION

1. Al + Fe3O4 Al2O3 + Fe



3. CH3OH + O2 CO2 + H2O


4. P4O10 + H2O H3PO4


5. PCl5 + H2O H3PO4 + HCl


6. SbCl5 + H2O SbOCl3 + HCl


7. MgO + Si Mg + SiO2


8. CaCl2 + Na2CO3 CaCo3 + NaCl


9. C6H6 + O2 CO2 + H2O


10. Al2S3 + H2O Al(OH)3 + H2S


11. C2H6 + O2 CO2 + H2O


12. KClO3 KCl + KClO4


13. KBr + Cl KCl + Br2


14. (NH4)2SO4 + NaOH NH3 + H2O + Na2SO4


15. Calcium Phosphate reacts with Sulfuric Acid to give Phosphic acid and Calcium Sulfate as a solid. Writ the balanced equation for this reaction.
ANSWERS



1.


2.
2. 8Al + 3Fe3O4 4Al2O3 + 9Fe


3. 2CH3OH + 3O2 2CO2 + 4H2O

Or CH3OH + 1 ½ O2 à CO2 + 2H2O

3. P4O10 + 6H2O 4H3PO4


4. PCl5 + 4H2O H3PO4 + 5HCl


5. SbCl5 + H2O SbOCl3 + 2HCl


6. 2MgO + Si 2Mg + SiO2


7. CaCl2 + Na2CO3 CaCo3 + 2NaCl


8. 2C6H6 + 15O2 12CO2 + 6H2O

Or C6H6 + 7 ½ O2 à 6CO2 + 3H2O

9. Al2S3 + 6 H2O 2Al(OH)3 + 3H2S


10. 2C2H6 + 7O2 4CO2 + 6H2O

Or C2H6 + 3 ½ O2 à 2 CO2 + 3H2O

11. 4KClO3 KCl + 3KClO4


12. 2KBr + 2Cl 2KCl + Br2


13. (NH4)2SO4 + 2NaOH 2NH3 + 2H2O + Na2SO4


14. Ca3(PO4)2 + 3H2SO4 à 3CaSO4 + 2H3PO4

Unit Conversion

Dimensional Analysis (Factor Label Method)

1. 1.2 kg = ________dg

2. 2.00 x 10-5m = ________in

3. 25.4 mm = ________cm

4. 1.2 miles = ________km

5. 15.47 m³ = ________km³

6. 17.0 ft/s = ________m/min

7. 342 miles/hr = ________km/s

8. 45.1 yards = ________cm

9. 1.45 L = ________gallons

10. 4.100 g = ________mg

11. 1.2 kg/yard = ________lbs/m

12. 2.00ft³/min = ________L/hour

13. 145 ml = ________cm³

14. 6.51 miles = ________cm

ANSWERS

1. 1.2 x 10⁴ dg

2. 7.87 x 10 ^–4 in

3. 2.54 cm

4. 1.9 km

5. 1.547 x 10 ^–8 km³

6. 311 m.min

7. 0.153 km/s

8. 4.12 x 10³

9. 0.383 gallon

10. 4.100 x 10³ mg

11. 2.9 lbs/m

12. 3.40 x 10³ L/hr

13. 145 cm³

14. 1.05 x 10⁶ cm

MGCCC, Perk Learning Lab

Metric to Metric Conversions

1. 14.4 m = __________ cm

2. 564 cg = __________ g

3. 58dg = __________ mg

4. 800L = __________ kL

5. 0.0687 km = __________ mm

6. 51.0 hg = __________ g

7. 210 cL = __________ dL

8. 4.51 x 10³ F = __________ mL

9. 45700 cg = __________ kg

10. 24.6 kL = __________ FL

11. 82.4 nm = __________ Fm

ANSWERS

1. 1440cm

2. 5.64 g

3. 5800 mg

4. 0.800 kL

5. 68700 mm

6. 5100 g

7. 21.0 dL

8. 4.51 mL

9. 0.457 kg

10. 2.46 x 10 ¹⁰ FL

11. .0824 Fm

Metric to Metric Conversions

The essence of metric to metric conversions is recognizing the abbreviated base units.

Base Units: m (meter), L (liter), g (gram), s (seconds)

These are not all of the base units, but are the most common that you will encounter. Each base unit can have any of the prefixes listed in the chart below. As with any other prefix, the prefix means the same thing regardless of the abbreviated base unit it is attached to.

Giga (G)---------- it takes 10⁹ (billion) base units = 1 G

Mega (M)--------- it takes 10⁶ (million) base units = 1 M

---------------------

---------------------

Kilo (K)----------- it takes 1000 base units = 1 K

Hecto (H)--------- it takes 100 base units = 1 H

Decka (Da)------- it takes 10 base units = 1 Da

Base Units: m, L, g, s

Deci (d)----------- it takes 1 base unit = 10 deci

Centi (c)---------- it takes 1 base unit = 100 centi

Milli (m)---------- it takes 1 base unit = 1000 milli

---------------------

---------------------

Micro (m)--------- it takes 1 base unit = 10⁶ (million) micro

---------------------

---------------------

Nano (n)---------- it takes 1 base unit = 10⁹ (billion) nano

There are two ways to work metric to metric conversions. The first way is to do the math by using the numerical conversion relationships listed in the chart above.

Ex: Convert 35 mg to g.

? g = 35 mg x 1 g = 0.000035 g

1,000,000 mg

The other way is to count the lines in the chart (every line except the one you start on), start from your given information and counting each line until you get to what you want to convert.

***Move Up the chart: Move decimal to the LEFT

***Move DOWN the chart: Move decimal to the RIGHT (remember RIGHT DOWN to the nitty gritty)

Ex: Convert 35 mg to g.

Begin at prefix m (micro) and count UP to g (gram). You counted UP six places, therefore you move your decimal to the LEFT six places. If your decimal is not shown, it is understood to be at the end of the number.

35. mg = 0.000035 g

Ex: Convert 0.26 ML to mL

Begin at prefix M (mega) and count DOWN to prefix m (milli). You counted DOWN nine places, therefore, you move your decimal to the RIGHT nine places. So, you will add seven zeros.

0.26 ML = 260,000,000. mL

PRACTICE PROBLEMS (and solutions) ON BACK

PRACTICE PROBLEMS

1) 170.4 m =______________ cm

2) 564 Dag =______________ g

3) 58 dg =______________ mg

4) 600 L =______________ KL

5) 0.0923 Km =______________ mm

6) 49 Hg =______________ g

7) 210 cL =______________ dL

8) 4.51 x 10³ mL =______________ mL

9) 45700 cg =______________ Kg

10) 24.6 KL =______________ mL

11) 82.4 nm =______________ mm

ANSWERS

1) 17040cm

2) 5640 g

3) 5800 mg

4) 0.600 KL

5) 92300 mm

6) 4900 g

7) 21.0 dL

8) 4.51 mL; write out 4.51 x 10³ mL = 4510 mL, the decimal is understood to be after the zero. Now count UP from the m prefix to the m (milli) prefix. You counted three places, so move the decimal to the LEFT three places = 4.51 mL

9) 0.457 Kg

10) 2.46 x 10¹⁰ mL or 24,600,000,000 mL

11) 0.0824 mm

Molarity, Molality and Normailty

Molarity = Moles of solute/Liters of Solution (abbreviation = M)

Molality = Moles of solute/Kg of Solvent (abbreviation = m)

Normality = equivalent solute/Liters of Solution (abbreviation = N)

1. How many moles of ethyl alcohol, C₂H₅OH, are present in 65ml of a 1.5M solution?

2. How many liters of a 6.0M solution of acetic acid CH₃COOH, contain 0.0030 mol acetic acid?

3. You want 85 g of KOH. How many of 3.0 m solution of KOH will provide it?

4. If you dissolve 0.70 moles of HCl in enough water to prepare 250ml of solution, what is the molarity of the solution you have prepared?

5. A solution is prepared by adding 2.0L of 6.0 M HCl to 500 ml of a 9.0 M HCl solution. What is the molarity of the new solution? (Remember, the volumes are additive)

6. What is the molality of a 0.10M solution of ethylene glycol C₂H₆O₂? The solutions density is 0.90 g/ml.

7. Convert the following Molarities to Normalities.

a. 2.5M HCl = _________ N

b. 1.4M H₂SO₄ = _________ N

c. 2.0M PbCl₃ = _________N

d. 1.0M NaOH = ________N

e. 0.5M Ca(OH)₂ = __________N

8. A commonly purchased disinfectant is a 3.0% (by mass) solution of hydrogen peroxide (H₂O₂) in water. Assuming the density of the solution is 1.0g/cm³ , calculate the molarity, molality and mole fraction of H₂O₂.

9. A solution is made by dissolving 25g of NaCl in enough water to make 1.0L of solution. Assume the density of the solution is 1.0 g/cm³. Calculate the molarity, normality and molality of the solution.

10. When two solutions react and you have their concentrations expressed in normality, you can write: V_aN_a = V_bN_b. How many liters of a 0.30 N solution of KMnO₄ can react with 5.0 L of a 0.10 N solution of H₂C₂O₄ ?

11. What is the concentration (M) of NaOH if 75.6 ml of NaOH is required to react with 270.0ml of a 0.3M solution of H₂SO₄?

ANSWERS

1. 0.098 mol alcohol 8. 0.88M, 0.91m, mole fraction 1.6 x 10^-2

2. 5.0 x 10^-4 L solution 9. 0.43M, 0.43N, 0.44m

3. 5.9 x 10 ² g solution 10. 1.7L

4. 2.8M 11. 2.1M

5. 6.6M

6. 0.11m

7. a. 2.5N

b. 2.8N

c. 6.0N

d. 1.0 N

e. 1N

PERCENT BY MASS

1a. How many moles of solute are contained in 343 grams of a 23% aqueous solution of MgCr₂O₇?

1b.How many grams of solute are contained in the solution of question 1a?

1c. How many grams of water (the solvent) are contained in the solution of question 1a?

1d. How many molecules of water are contained in the solution of question 1a?

1e. How many Cr atoms are contained in the solution of question 1a?

2a. The density of a 0.13% solution of NaCl is 2.16 g/mL. What mass of NaCl would be required to prepare 45L of this solution?

2b. How many molecules of solute are contained in the solution of question 2a?

3a. Specific gravity of a 33% (NH₄)₂SO₄ solution is 1.1. What mass of (NH₄)₂SO₄ would be required to prepare 201mL of this solution?

3b. How many moles of solute are contained in the solution of question 3a?

ANSWERS

1a) ? g MgCr₂O₇ = 343g x 0.23 = 78.89 g MgCr₂O₇

? mol MgCr₂O₇ = 78.89 g MgCr₂O₇ x (1mol / 240g MgCr₂O₇ ) = 0.329 mol MgCr₂O₇

1b) ?g MgCr₂O₇ = 343g x 0.23 = 78.89 g MgCr₂O₇

1c) ? g H₂O = 343 g solution – 78.89g MgCr₂O₇ = 264.11 g H₂O

1d) ? moles H₂O = 264.11 g H₂O x (1 mol/18 g H₂O)=14.67 mol H₂O

? Molecules H₂O= 14.67mol H₂O x (6.02 x 10²³molecules/1mol H₂O)= 8.83 x 10²⁴ molecules H₂O

1e) ? molecules of MgCr₂O₇ = 0.329 mol MgCr₂O₇ x (6.02 x 10²³ molecules/ 1 mol)

= 1.98 x 10²³ molecules of MgCr₂O₇

? atoms of Cr = 1.98 x 10²³ molecules of MgCr₂O₇ x ( 2 Cr atoms/ 1 molecule) = 3.96 x 10²³ Cr atoms

2a) ? g NaCl impure = (2.16 g NaCl /1 mL) x (1000 mL / 1 L) = 2160 g/L impure NaCl

?g NaCl pure= (2160 g NaCl/ L) x 0.0013 = 2.808 g / L NaCl pure

? g NaCl = 45 L x (2.808 g NaCl/ 1 L) = 126.36 g NaCl

2b) ? moles NaCl = 126.36 g NaCl x (1mol / 58g NaCl)= 2.17 mol NaCl

? molecules NaCl = 2.17 mol NaCl x (6.02 x 10²³ molecules/ 1 mol) = 1.31 x 10²⁴ molecules NaCl

3a) specific gravity = density

? g (NH₄)₂SO₄ impure = (1.1g (NH₄)₂SO₄/ 1 mL) x (1000mL / 1L) = 1100g (NH₄)₂SO₄/ L impure

? g (NH₄)₂SO₄ pure= (1100 g (NH₄)₂SO₄ / L)(0.33) = 363 g (NH₄)₂SO₄ /L pure

? g (NH₄)₂SO₄ = 0.201L x (363g (NH₄)₂SO₄ / L) = 72.96 g (NH₄)₂SO₄

3b) ? moles (NH₄)₂SO₄ = 72.96 g (NH₄)₂SO₄ x (1 mol/ 132g (NH₄)₂SO₄) = 0.55 mol (NH₄)₂SO₄

DENSITY

Practice Problems and Answers

MGCCC, Perk Learning Lab

TLM

1. What is the density of benzene in grams per milliliter if 5.5L of benzene weigh 4.8kg?

2. The density of air is 1.3 g/L. How many liters of air weigh 2.0 lb?

3. Given a cylinder with diameter of 39 mm and height of 39 mm, and mass of 1.0 Kg, what is it’s density?

4. How many ounces does a 250ml sample of sulfuric acid weigh? (density = 1.30 g/ml)

5. What is the density of the Earth if 1.00 yd³ weighs 22.0 lb? Express your answer in lb/ft³.

6. The density of brass is 8.0 g/ml. A cube of brass 2.0 inches on an edge will weigh how many grams?

7. An unknown substance has a volume of 0.114 L and has a mass of 467g. What is its density in g/mL?

8. The density of Ethanoic Acetic Acid is 1.05 g/cm³. What is the mass in kilograms of a volume of 5L?

ANSWERS

1) D = mass ?D= 4.8Kg x 1000g x 1L = 0.87 g/mL

vol 5.5L 1Kg 1000mL

2) V= m ?g = 2 lb x 453.6 g = 907.2 g

D 1 lb

?V = 907.2g = 907.2g x 1L = 697.8 L

1.3g 1.3g

1L

3)Volume of a cylinder v= P r²h

r = ½ d, r = ½ (39mm) = 19.5 mm

v= P (19.5 mm)²(39mm) = 46589 mm³

? cm³ = 46589 mm³ x (1m)³ x (100cm)³ = 46.5 cm³

(1000mm)³ (1m)³

D = 1 Kg x 1000g = 21.5 g/cm³

46.5 cm³ 1Kg

4) m= vD

?oz = 250ml x 1.3g x 1 oz = 11.5 oz

1 mL 28.35 g

5) ? lb = 22 lb x (1yd)³ = 0.815 lb/ft³

ft³ 1 yd³ (3ft)³

6) Volume of a cube v = s³

? v = (2in)³ x (2.54 cm)³ = 131cm³

(1 in)³

? v = 131 cm³ x 1 mL = 131 ml

1 cm³

m= vD = 131 mL x 8g = 1048g

1 mL

7) D = m = 467g x 1 L = 0.467 g/mL

v 0.114L 1000mL

8) ? v = 5 L x 1000mL x 1 cm³ = 5000cm³

1L 1mL

?m = vD= 5000cm³ x 1.05g x 1 Kg = 5.25 Kg

1 cm³ 1000g

Isotope Abundance and Atomic Weight

Ex: The natural abundance for boron isotopes is: 19.9% ¹⁰B (10.013 amu) and 80.1% ¹¹B (11.009amu).

Calculate the atomic weight of boron.

Aw = [(%abundance of isotope) (mass of isotope)] + [(%abundance of isotope) (mass of isotope)] + [….]

Aw = the atomic weight found in the periodic table

% abundance of EACH isotope is converted to decimal ex: 19.9% ÷ 100% = 0.199

The equation continues on[….] based on the number of isotopes in the problem.

Aw = [(0.199)(10.013)] + [(0.801)(11.009)]

Aw = [1.992587] + [8.818209]

Aw = 10.810796 so, the atomic weight of B = 10.811

If you look in the periodic table you will be able to check that our answer is correct!

The atomic mass of lithium is 6.94, the naturally occurring isotopes are ⁶Li = 6.015121 amu, and ⁷Li = 7.016003 amu.

Determine the percent abundance of each isotope.

Aw = [(%abundance of isotope) (mass of isotope)] + [(%abundance of isotope) (mass of isotope)] + [….]

6.94 =[(% ⁶Li)(6.015121)] + [(%⁷Li)(7.016003)]

Since I don’t know what the percentage are, I will have to use variables.

100% of Lithium is determined by these two naturally occurring isotopes.

We will let ⁶Li = x and ⁷ Li = 1-x; we use 1 – x instead of 100 – x because the small number is easier to work with.

(in other words we reduced 100% to decimal form 1.00)

Now let’s plug our variables in: 6.94 =[(% ⁶Li)(6.015121)] + [(%⁷Li)(7.016003)]

6.94 = [(x)(6.015121)] +[(1-x)(7.016003)]

6.94 = 6.015121x + 7.016003 – 7.016003x

Combine like terms: 6.94 -7.016003 = (6.015121x - 7.016003x)

-0.076003 = -1.000882 x

Solve for x: -0.076003 = x

-1.000882

X = 0.075936, therefore ⁶Li = 0.075936 x 100% = 7.59%

1-X = 1 -0.075936 = 0.924064, therefore ⁷Li = 0.924064 x 100% = 92.41%

PROBLEMS (SOLUTIONS ON BACK)

1. Calculate the weight of silicon using the following data for the percent natural abundance and mass of each isotope:

92.23% ²⁸Si (27.9769 amu); 4.67% ²⁹ Si (28.9765); 3.10% ³⁰Si (29.9738 amu).

2. Thallium has two stable isotopes, ²⁰³Tl and ²⁰⁵Tl. Knowing that the atomic weight of thallium is 204.4,

which isotope is the more abundant of the two?

3. Verify that the atomic mass of magnesium is 24.31, given the following:

24Mg= 23.985042amu, 78.99%

25Mg= 24.985837 amu, 10.00%

26Mg= 25.982593, 11.01%

4. Copper exists as two isotopes: ⁶³Cu (62.9298 amu) and ⁶⁵Cu (64.9278 amu).

What are the percent abundances of the isotopes?

SOLUTIONS

Aw = [(%abundance of isotope) (mass of isotope)] + [(%abundance of isotope) (mass of isotope)] + [….]

1. Aw= [(0.9223)(27.9769)] + [(0.0467)(28.97650)] + [(0.031) (29.9738)]

Aw = 25.803 + 1.353 + 0.929

Aw for Silicon = 28.085

Hint: double check the periodic table to see if the answer is the same or close

2. ²⁰³Tl % = x

²⁰⁵Tl % = 1 – x

204.4 = [(x)(203)] + [(1- x)(205)]

204.4 = 203x + 205 – 205x

-0.6 = -2x

-0.6 = x

-2

X= 0.3 ²⁰³Tl = 0.3 x 100% = 30 %

²⁰⁵Tl = 1 –x = 1 – 0.3 = 0.7 x 100% = 70%

Therefore ²⁰⁵Tl = 70% is more abundant.

3. Aw= [(0.7899)(23.985042)] + [(0.1)(24.985837)] + [(0.1101)(25.982593)]

Aw = 18.946 + 2.499 + 2.861

Aw for Mg = 24.31, this checks with the Aw that was given in problem #3.

4. Since the overall atomic weight for copper is not given in the problem, you must look it up in the

periodic table to work this solution. Aw for Cu = 63.546

⁶³Cu % = x

⁶⁵Cu % = 1 – x

63.546 = [(x)(62.9298)] + [(1-x)(64.9278)]

63.546 = 62.9298x + 64.9278 – 64.9278x

-1.3818 = -1.998x

-1.3818 = x

-1.998

X = 0.6916 ⁶³Cu = 0.6916 x 100% = 69.16%

⁶⁵Cu = 1 – x = 1 – 0.6916 = 0.3084 x 100% = 30.84%