Stochiometry

Question 1

What is the empirical formula of a compound containing 40.0% sulfur and 60.0% oxygen by mass?

Answer 1

SO₃. (40 grams of Sulfur divided by 32 = 1.25 and 60 grams of Oxygen divided by 16 = 3.75)

Question 2

A compound is found to contain 23.3% magnesium, 30.7% sulfur and 46.0% oxygen. What is the empirical formula of this compound?

Answer 2

• 23.3g/24 = 0.97 Mg
• 30.7/32 = 0.96 S
• 46/16 = 2.88 O
• Thus: MgSO₃

Question 3

Difference between molarity and molality

Answer 3

• Molarity = M = mol/L
• molality = m = mol/kg

Question 4

Description of composition by % mass

Answer 4

%mass = mass of species of interest / total mass * 100

Question 5

Balance the combustion of propanol:

C₃H₈O + O₂ → CO₂ + H₂O

Answer 5

2C₃H₈O + 9O₂ → 6CO₂ + 8H₂O

Question 6

Balance:

K₂Cr₂O₇ (aq) + HCl (aq) → KCl (aq) + CrCl₃ (aq) + H₂O (l) + Cl₂ (g)

Answer 6

K₂Cr₂O₇ (aq) + 14 HCl (aq) →

2CrCl₃ (aq)+ 7H₂O (l) + 3Cl₂ (g) + 2KCl (aq)

Question 7

How do you balance red-ox reactions?

Answer 7

1. Balance elements in the equation other than O and H.
2. Balance the oxygen atoms by adding the appropriate number of water (H2O) molecules to the opposite side of the equation.
3. Balance the hydrogen atoms (including those added in step 2 to balance the oxygen atom) by adding H+ ions to the opposite side of the equation.
4. Add up the charges on each side. Make them equal by adding enough electrons (e-) to the more positive side. (Rule of thumb: e- and H+ are almost always on the same side.)
5. The e- on each side must be made equal; if they are not equal, they must be multiplied by appropriate integers (the lowest common multiple) to be made the same.
6. The half-equations are added together, canceling out the electrons to form one balanced equation. Common terms should also be canceled out.

Question 8

If an equation is being balanced in basic conditions, what last step do you have to do?

Answer 8

If the equation is being balanced in a basic solution, through the addition of one more step, the appropriate number of OH- must be added to turn the remaining H+ into water molecules.

Question 9

What is the limiting reagent of the given reaction?

3Xox + Ared → 3Xred + Aox

Given: You use 60 grams of Xox and 63 grams of Ared

Given: the molecular weight of Xox is 2 amu, and Ared is 7 amu.

Answer 9

• The first thing you do is convert everything in moles
• 1 amu = 1 g/mol.
• Xox: 60 g / 2 amu = 30 mols.
• Ared: 63 g / 7 amu = 9 mols.
• Now here’s where stoichiometry comes in: divide the mols by the stoichiometric coefficient of the species:
• 30 mols / 3 = 10 for Xox
• 9 mols / 1 = 9 for Ared
• Now compare the values. 9 is the smallest, so Ared is the limiting reactant.

Question 10

What is the theoretical yield of the given reaction?

3Xox + Ared → 3Xred + Aox

Given: You use 60 grams of Xox and 63 grams of Ared

Given: the molecular weight of Xox is 2 amu, and Ared is 7 amu and Xred is 10 amu.

Answer 10

First, find who’s the limiting reagent.

9 mols of Ared * 3 mols of Xred per 1 mol of Ared = 27 mols.

Lastly, convert mols to grams: 27 mols * 10 g/mol = 270 g

Question 11

1) Sox + 5Xred → 3Sred + 3Xox
2) 3Xox + Ared(limiting reagent) → 3Xred + Aox
3) Xox(left over) + 2I- → 2Xred + I2
4) I2 + 2Tred → 2I- + Tox

after a long time doing drip by drip titration, you finally saw the dark color change to colorless. You noted down the initial and final volume reading of your pippette to be 300 mL and 200 mL, respectively. The concentration of the titrant you used was 10 M. You dissolved 1/2 mols of the standard to begin with. How much analyte was there?

Answer 11

• First, convert everything to mols (amount). n = MV. For the titrant (T_red) it is 10 M x (0.3 L - 0.2 L) = 1 mol
• For the standard (S_ox), it is already given to you in mols. However, if it’s not, you have to convert it to mols.
• We know from the notes above that X_ox - X_ox(left over) = the amount of analyte, after taking into account of stochiometric ratios.
• Here are the stochiometric ratios:
  
  • From step 4
    I₂ : 2T_red
  • From step 3
    X_ox(left over) : I₂
  • From step 2
    3X_ox : A_red(limiting reagent)
  • From step 1
    S_ox : 3X_ox
• X_ox = 0.5 mol S_ox * 3X_ox / S_ox = 1.5 mol X_ox
• X_ox(left over) = 1 mol T_red * I₂ / 2T_red * X_ox(left over) / I₂ = 0.5 mol X_ox(left over)
• For every A_red(limiting reagent), you eat up 3 X_ox, thus:
  X_ox - 3A_red(limiting reagent) = X_ox(left over)
  1.5 - 3 * A_red(limiting reagent) = 0.5
  A_red(limiting reagent) = 1/3 mol