Unit 3 - Quantity of chemical formulas

$N_A=n=6.022\cdot 10^{23}$

$N=\text{number of particles}$

$n=\text{number of moles}$

$N_A=\text{Avagadro constant}$

$N=n{N}_A-\text{The General Formula}$

$\text{Mass}=M=\text{Number of moles}\cdot \text{Molar Mass}$

$$\frac{g}{M|n}$$

Molar mass is the mass in grams of 1 mole of a substance
Unit:$g/mol$

• For Elements: Molar mass = relative atomic mass
• For Molecules: Molar mass = Molecular mass
  • Molecular mass is the mass of one molecule of a covalent compound.

Percent Composition

$\%composition=\frac{\text{mass of element in compound}\cdot 100\%}{\text{total mass of compound}}$

zinc = 67.10%
sulpher 32.90%

Dimensional Analysis

$$\text{Given unit}\times \frac{\text{desired unit}}{\text{given unit}}=\text{desired unit}$$

Determining to formula of a hydrate

22.8g water = 45.6% water

1. Molar mass of Ba(OH)₂ = 137.33 (Ba) + 2 * 16 (O) + 2 * 1.01 (H) = 171.35 g/mol
2. Moles of Ba(OH)₂ = mass / molar mass = 27.2g / 171.35 g/mol = 0.1587 mol
3. Moles of N2O = Mass / molar mass = 22.8g / 18.02g/mol = 1.26526082 mol
4. Mole ratio = moles of H₂O / moles of Ba(OH)₂ = 1.265 mol / 0.1587 mol = 7.97 ≈ 8

Limiting Reagents

The reactant that determines how much product will be formed

• Exact stoichiometric amounts in chemicals are rare and impractical
  • In practice, there are limiting effects are not perfectly consistant
• The reactant that gets used up first is called the Limiting reagent. You can only have one.
  • Remaining reactants are called excess reagents

Homework

Mole-calculation pg. 1 - 2024-11-05

a) 1.2 * 10^24
b) 6.02E+24
c) 9.0E+23
d) 2.5E+23
e) 1.3E+21

2
a) 0.102
b) 5.00E+2
c) 8.6
d) 1.4E-3

a) 6.48
b) 3.4
c) 8.49x10^24
d) 1.19x10^-1 x 2 x$N_A$=
e) 4.26 x 10

PART TWO
a) 225 / 180.175 = 1.24878590259 = 1.25
b) 0.417 mol
c) 0.42 mol
d) 3.88mol
e) 1.46 molga weight
f) 1450 g / 196.96657 (au weight) = 7.36165533065 = 7.36

PART THREE
a) 4.002602 * 12.4 = 49.6322648 = 49.6 g
b) 15.1g
c) 4344b
d) 105g

2024-11-07 practice

1. 1.07 mol
2. 124 mol
3. 14.2 mol
   p2
4. 0.652 mol
5. 0.156 mol
6. 17.6129134 mol (17.6 mol)

1.5x10

1. Do 1.0 mol samples of different compounds all have the same mass? Explain your answer.

No, 1.0 mol samples of different compounds do not have the same mass. The mass of a sample depends on the molar mass of the compound, which varies depending on the elements in the compound and their quantities. Molar mass is the mass of 1.0 mol of a substance, and it differs from one compound to another.

2. Describe how chemists use mass as a way of getting a precise estimate of the number of entities in a sample.

Chemists use the mass of a sample and its molar mass to estimate the number of entities (atoms, molecules, or ions) in that sample. By dividing the mass of the sample by the molar mass, they can calculate the number of moles. Then, using Avogadro’s number (6.022 x 10²³ entities per mole), they can determine the number of individual entities in the sample.

3. Why is an amount, obtained using the equation n = always an estimate rather than an exact value?

The amount calculated using the equation$n=\frac{\text{mass}}{\text{molar mass}}$is an estimate because the values used for mass and molar mass are subject to measurement uncertainty. The molar mass of a compound is an average, and the mass of a sample is typically measured with some degree of precision but not perfect accuracy.

4. Calculate the molar mass of each of the following compounds:

(a) Iron(III) oxide, Fe₂O₃ (in rust)
Molar mass of Fe₂O₃ =$2\times 55.85\text{g/mol (Fe)}+3\times 16.00\text{g/mol (O)}$
=$111.7\text{g/mol}+48.00\text{g/mol}$
= 159.7 g/mol

(b) Calcium carbonate, CaCO₃ (in blackboard chalk)
Molar mass of CaCO₃ =$1\times 40.08\text{g/mol (Ca)}+1\times 12.01\text{g/mol (C)}+3\times 16.00\text{g/mol (O)}$
=$40.08\text{g/mol}+12.01\text{g/mol}+48.00\text{g/mol}$
= 100.09 g/mol

(c) Octane, C₈H₁₈ (in gasoline)
Molar mass of C₈H₁₈ =$8\times 12.01\text{g/mol (C)}+18\times 1.008\text{g/mol (H)}$
=$96.08\text{g/mol}+18.14\text{g/mol}$
= 114.22 g/mol

(d) Calcium chlorate, Ca(ClO₃)₂ (in fireworks)
Molar mass of Ca(ClO₃)₂ =$1\times 40.08\text{g/mol (Ca)}+2\times \left(1\times 35.45\text{g/mol (Cl)}+3\times 16.00\text{g/mol (O)}\right)$
=$40.08\text{g/mol}+2\times (35.45\text{g/mol}+48.00\text{g/mol})$
=$40.08\text{g/mol}+2\times 83.45\text{g/mol}$
=$40.08\text{g/mol}+166.90\text{g/mol}$
= 206.98 g/mol

(e) Ammonium carbonate, (NH₄)₂CO₃ (in smelling salts)
Molar mass of (NH₄)₂CO₃ =$2\times \left(1\times 14.01\text{g/mol (N)}+4\times 1.008\text{g/mol (H)}\right)+1\times 12.01\text{g/mol (C)}+3\times 16.00\text{g/mol (O)}$
=$2\times (14.01\text{g/mol}+4.032\text{g/mol})+12.01\text{g/mol}+48.00\text{g/mol}$
=$2\times 18.042\text{g/mol}+12.01\text{g/mol}+48.00\text{g/mol}$
=$36.084\text{g/mol}+12.01\text{g/mol}+48.00\text{g/mol}$
= 96.094 g/mol

5. Calcium chloride, CaCl₂, is used as a drying agent to protect electronics during shipping. Calculate the amount of calcium chloride in a 10.0 g sample.

To calculate the number of moles, use the molar mass of CaCl₂:
Molar mass of CaCl₂ =$1\times 40.08\text{g/mol (Ca)}+2\times 35.45\text{g/mol (Cl)}$
=$40.08\text{g/mol}+70.90\text{g/mol}$
= 110.98 g/mol

Number of moles:
$n=\frac{\text{mass}}{\text{molar mass}}=\frac{10.0\text{g}}{110.98\text{g/mol}}=0.0902\text{mol}$

6. A typical energy drink contains 80.0 mg of the stimulant caffeine, C₈H₁₀N₄O₂. Calculate the amount of caffeine in the energy drink.

Convert mass to grams:
$80.0\text{mg}=0.0800\text{g}$

Molar mass of caffeine:
Molar mass of C₈H₁₀N₄O₂ =$8\times 12.01\text{g/mol (C)}+10\times 1.008\text{g/mol (H)}+4\times 14.01\text{g/mol (N)}+2\times 16.00\text{g/mol (O)}$
=$96.08\text{g/mol}+10.08\text{g/mol}+56.04\text{g/mol}+32.00\text{g/mol}$
= 194.20 g/mol

Number of moles of caffeine:
$n=\frac{\text{mass}}{\text{molar mass}}=\frac{0.0800\text{g}}{194.20\text{g/mol}}=4.12\times 10^{-4}\text{mol}$

7. A kernel of popping corn contains 1.22 x 10⁻³ mol of water. What is the mass of this water?

Molar mass of water (H₂O) =$2\times 1.008\text{g/mol (H)}+16.00\text{g/mol (O)}$
=$2.016\text{g/mol}+16.00\text{g/mol}$
= 18.016 g/mol

Mass of water:
$\text{mass}=n\times \text{molar mass}=1.22\times 10^{-3}\text{mol}\times 18.016\text{g/mol}$
= 0.022$\text{g}$

2024-11-11 pg 288 1-8

1. Total weight: 50g
   1. Calcium ($20/50$) = $0.400\ast 100=40\%$
   2. Carbon (6/50) = $0.120\times 100=12\%$
   3. Oxygen (24/50) = 0.480 * 100 = 48%
2. Total weight: 5g
   1. Carbon: $\frac{2.61}{5.0}\times 100=52\%$
   2. Hydrogen: $\frac{0.66}{5}\times 100=13\%$
   3. Oxygen: $100-(52+13)=35\%$
3. The law of definite proportions states that a compound will always have the same proportions of it’s elements/subcompounds as if it didn’t, it would not be the same compound. So if one H2O2 has 2 Oxygen. 2 H2O2 has 4 oxygen, hence double (answer should have more clearly specified ratios/by mass)
4. 1. a) benzene: $C_6{H}_6$ $\text{Mass of hydrogen}=10.00g-9.20g=0.800g$
   2. b) Carbon: $\frac{9.2}{10}=92\%$ hydrogen: $\frac{0.8}{10}=8\%$
5. a. given the total ratio of carbon to hydrogen in both pairs combined, you get a 5:8 ratio in a which is the highest compared to the 5:10 in b and 4:10 in c
   1. You can easily figure this out without mass calculations using ratios as all compounds are hydrocarbons
   2. a) 2
   3. b) tied at $85.7\%$
   4. c) 1
6. (a) Decreases due to CO2 gas escape. (b) Increases as lighter oxygen atoms are removed.
7. 1. H₃PO₄ | H: 3.09%, P: 31.60%, O: 65.31%
   2. Cu₂S | Cu: 66.7%, S: 33.3%
   3. Fe₂O₃ | Fe: 69.9%, O: 30.1%
   4. B(OH)3 | B: 17.5%, H: 4.9%, O: 77.6%
8. I predict that Ammonium nitrate will have a higher percentage as there is only a high amount of (light) Hydrogen and 3 oxygens compared to the 12 hydrogens, 4 oxygens and a phosphorus in Ammonium phosphate
   1. Ammonium nitrate has a higher percentage of nitrogen (35%) compared to ammonium phosphate (28.21%).
9. Gold fingerprinting

2024-11-18 - Empical / molecular formula

1. (30.031 / 150g/mol = 5) * CH2O = C5H10O5 (Empirical to molecular)

2024-11-20

Given 25.0g of aluminum how much Al2O3 is Produced
4Al+3 O2 → 2 Al2O3

AL: 26.98
O: 16.00

47.2386953293g → 47.2g

2024-11-21

Mass to mass calculations

First

2 N2H4 + 1 N2O4 → 3 N2 + 4 H2O

Di nitrogen: 28.0134 g/mol
hydrazine: 32.0452 g/mol
50g/32.0452 g/mol = 1.56029608178 mol

1.56029608178 mol * $\frac{3}{2}$ = 2.34044412267 mol (of nitrogen)
2.34044412267 * 28.0134 = 65.5637974 = 65.6g (65.5 if using periodic table numbers)

Second

2H2 + O2 = 2H2O

H2: 10
O2: 7

a) hydrogen. Oxygen is exaxx
b) there ar e two leftoven molecules of O2
c) 10

Third

1 Li3N + 3H2O +1 HN3 +3LiOH

4.87g of LiN + 5.80g of H2O

4.87 / 34.83 g/mol = 0.1398 mol
Water: 5.80 / 18.02 g/mol = 0.32186 mol * 1 mol Li3N/3 mol 3H2O = 0.1062881 “units”

Fourth

P4 + 5O2 → P4O10
1.00g of P4 + 2.60 * 10^23 → x P4O10

2.6e23/6.022e23 = 0.43175024908 mol Oxygen / ( 1 mol P4 /5 mol O2 ) → 0.0863500498
1/124 = 0.00806451612 mol Phosphorus (limiting reagent)

0.00806451612 * 283.889 = 2.28942741679g of P4H10 = 2.29g

2024-11-27

To determine the mass of ammonium nitrate (NH₄NO₃) produced from 4.95 g of ammonia (NH₃) with 89.5% efficiency, we can follow these steps:

Balanced: HNO3 + NH3 → NH4NO3

ratio: 1:1 ammonia to ammonium nitrate

Molar mass of NH3: 17.031g/mol
4.95g / 17.031g/mol = 0.2906464682 mol

Molar mass NH4NO3: 80.043g/mol

0.2906464682 mol * 80.043g/mol = 23.2642152541g

efficency

23.2642152541g * 0.895 = 20.8214726524 = 20.8g
20.8g

To calculate the mass of sulfur (S) required to produce 50.0 g of hydrogen sulfide (H₂S) in the reaction, we need to consider the given reaction

CH₄(g) + 4S(g) → CS₂(g) + 2H₂S(g)

4S:H2S = 4:2 → 2:1 ratio of sulfur to hydrogen sulfide in this equation

M of S = 32.065
M of H2S = 34.082

Moles of H2S needed:
50.0/34.082 = 1.4670500557 moles

1.4670500557 * 2/1 = 2.9341001114 mol

2.9341001114 mol * 32.065 = 94.0819200720g

(given value) 94.0819200720g / 0.752% = 125.1089362660g = 125g

Therefore, you need 125g of sulfer to make 50g of H2S in the that reaction