Formulas

Formation and Characterization

Percent Composition

The Percent Composition of a Compound

Percent composition = the percent by mass of each element in the compound

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• The relative amounts of the elements in a compound are expressed this way
• Consists of a percent value for each different element in the compound

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K₂CrO₇ = 40.3% K, 32.9 % O, & 26.8% Cr

K₂Cr₂O₇ = 26.5 % K, 38.1 % O, & 35.4 % Cr

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The percent by mass of an element in a compound is the number of grams of the element divided by the mass in grams of the compound multiplied by 100%.

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Practice

A compound is formed when 9.03 g Mg combines completely with 3.48 g N. What is the percent composition of this compound?

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72.2 % Mg, 27.8 % N

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Practice

When a 14.2 g sample of mercury (II) oxide is decomposed into its elements by heating, 13.2 g Hg is obtained. What is the percent composition of the compound?

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7.0 % O, 92.0% Hg

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Practice

Calculate the percent composition of these compounds.

a. ethane (C₂H₆)

80.0% C, 20.0 % H

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b. sodium hydrogen sulfate (NaHSO₄)

0.83 % H, 26.7 % S, 53.3 % O

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Practice

Calculate the percent of nitrogen in these common fertilizers

a. NH₃

82.4 % N

b. NH₄NO₃

35.0 % N

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Practice

Calculate the percent composition of the compound that forms when 222.6 g N combines completely with 77.4 g O.

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74.2 % N, 25.8% O

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Empirical and Molecular Formulas

Percent Composition as a Conversion Factor

o You can use percent composition to calculate the number of grams of an element

o You can say that the percentage is just out of 100 g of that compound

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Let’s say we want to find the amount of carbon/hydrogen in 82 g of C₃H₈

**So in a 100 g sample we would have 81.8 g C and 18.2 g H

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** remember the sum of the two masses should equal the total mass of the compound**

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Empirical Formulas

- The empirical formula of a compound shows the smallest whole number ratio of the atoms in the compound

o This means that it may or may not be the same as the molecular formula

o Ex: the empirical formula for hydrogen peroxide is HO but the molecular formula is H₂O₂

o Ex: the empirical and molecular formula for carbon monoxide is CO

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Empirical Formulas

-Gives the lowest whole-number ratio of the atoms of the elements in a compound.

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- Multiplying by different formulas could produce the formulas for different compounds

o Ex: ethyne, aka = acetylene (C₂H₂)= a gas used in welding torches

o Ex: styrene (C₈H₈) = main component in Styrofoam

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You can use the percent composition to find the empirical formula of a compound.

Example:

A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula of the compound?

Knowns:

o Percent of nitrogen = 25.9 % N

o Percent of oxygen = 74.1 % O

Unknowns:

o Empirical formula: N_?O_?

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*remember the subscripts are a molar ratio** so we’re done the empirical formula is N₂O₅

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Practice

Calculate the empirical formula of each compound.

a. 94.1% O, 5.9% H

HO

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b. 67.6 % Hg, 10.8% S, 21.6%O

HgSO₄

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Practice

1,6- diaminohexane is used to make nylon. What is the empirical formula of this compound if it is 62.1 % C, 13.8 %H, and 24.1% N?

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C₃H₈N

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Molecular Formulas

-The molecular formula of a compound is either the same as its experimentally determined empirical formula, or it is a simple whole-number multiple of its empirical formula.

-Once you know the empirical formula, you can use the molar mass of the compound to figure out what multiple the molecular formula is compared to the empirical formula

o Chemists use an instrument called a mass spectrometer to determine molar mass

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Molecular Formula

- Divide the molar mass of the compound by the empirical formula mass to determine the ratio

-Ex: the empirical formula for hydrogen peroxide is HO, giving it an empirical formula mass of 17.0 g/mol. We could experimentally determine the molar mass to be 34.0 g/mol

34.0 g/mol/17.0 g/mol = 2

o We would take this number multiplied by all the subscripts to obtain the molecular formula

₂ x(HO) = H₂O₂ : hydrogen peroxide

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Example (continued next page)

Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH₄N?

Knowns:

o Empirical formula : CH₄N

o Molar mass = 60.0 g/mol

Unknowns:

o Molecular formula

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Example

Empirical formula mass = 30.0 g/mol

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Molecular formula = C₂H₈N₂

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Practice

Find the molecular formula of ethylene glycol, which is used as antifreeze. The molar mass is 62 g/mol and the empirical formula is CH₃O.

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C₂H₆O₂

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Practice

Which pair of molecules has the same empirical formula?

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a. C₂H₄O₂, C₆H₁₂O₆

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b. NaCrO₄, Na₂Cr₂O₇

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Practice

Calculate the percent composition of calcium acetate → Ca(C₂H₃O₂)₂

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25.4 % Ca, 30.4 % C, 3.8 % H, 40.5 % O

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Practice

The compound methyl butanoate smells like apples. Its percent composition is 58.8% C, 9.8% H, and 31.4 % O and its molar mass is 102 g/mol. What is its empirical formula? What is its molecular formula?

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Empirical and molecular formula are both C₅H₁₀O₂

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Practice

Which of the following molecular formulas are also empirical formulas?

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a. ribose : C₅H₁₀O₅

b. ethyl butyrate : C₆H₁₂O₂

c. chlorophyll: C₅₅H₇₂MgN₄O₅

d. DEET: C₁₂H₁₇ON

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LIMITING REAGENT AND PERCENT YIELD�

Limiting and Excess Reagents

• In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms.

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• There isn’t always going to be the exact same number of reactants and products,
• Usually there will be more than enough of one reactant and a certain amount of another reactant

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Limited and Excess Reagents

• When determining which reactant is the limiting reactant, first convert all reactants into moles, at which time the limiting reactant can be identified

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• The amount of product formed can be found using the amount of the limiting reactant
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• You’ll want to make sure you understand this process… it is easy to lose track for this...

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Determining the Limiting Reagent in a Reaction

Example:

• Copper reacts with sulfur to form copper (I) sulfide according to the following chemical equation:

 2Cu(s) + S(s) → Cu₂S(s)

• What is the limiting reactant when 80.0 g Cu reacts with 25.0 g S?

 Known:

Mass of copper = 80.0 g

Mass of sulfur = 25.0 g

 Unknown:

Limiting reagent= ?

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Practice:

1. The equation for the complete combustion of ethane (C₂H₄) is:

C₂H₄ (g) + 3O₂ (g) →2CO₂(g) + 2H₂O(g)

If 2.70 mol C₂H₄ is reacted with 6.30 mol O₂, identify the limiting reactant.

O₂ is the limiting reactant

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Practice:

• 2. Hydrogen gas can be produced by the reaction of magnesium metal with hydrochloric acid.

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Mg(s) + 2HCl (aq) → MgCl₂(aq) + H₂(g)

• Identify the limiting reagent when 6.00 g HCl reacts with 5.00 g Mg.

• HCl is the limiting reactant

Using a Limiting Reagent to Find the Quantity of a Product

Example:

• What is the maximum number of grams of Cu₂S that can be formed when 80.0 g Cu reacts with 25 g S?

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• 2Cu(s) + S(s) → Cu₂S(s)

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Known:

• Limiting reagent = 1.26 mol Cu (from the other example problem)
• 1 mol Cu2S = 159.1 g Cu₂S (molar mass)

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Unknown:

• Mass of copper (I) sulfide = ? g Cu₂S

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Practice:

1. The equation for the incomplete combustion of ethane (C₂H₄) is:

C₂H₄ (g) + 2O₂ (g) →2CO (g) + 2H₂O(g)

 If 2.70 mol C₂H₄ is reacted with 6.30 mol O₂

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a. identify the limiting reagent

b. calculate the moles of water produced

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a. 5.40 mol O₂ required; C₂H₄ is the limiting reactant.

b. 5.40 mol H₂O

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Practice:

2. The heat from an acetylene torch is produced by burning acetylene (C₂H₂) in the presence of oxygen.

2C₂H₂ (g) + 5O₂ (g) → 4CO₂(g) + 2H₂O (g)

How many grams of water can be produced by the reaction of 2.40 mol C₂H₂ with 7.40 mol O₂?

43.2 g H₂O

Percent Yield

• Theoretical yield = the maximum amount of product that could be formed from given amounts of reactants

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• Actual yield = the amount of product that actually forms when the reaction is carried out in the laboratory

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Percent Yield

Percent Yield

• The percent yield is a measure of the efficiency of a reaction carried out in the laboratory
• Because the actual yield of a chemical reaction is often less than the theoretical yield, the percent yield is often less than 100%

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• Should not be larger than 100%

Calculating the Theoretical Yield of a Reaction

Example:

• Calcium carbonate, which is found in seashells, is decomposed by heating. The balanced equation for this reaction is:

CaCO₃(s) ^∆ CaO (s) + CO₂ (g)

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• What is the theoretical yield of CaO if 24.8 g CaCO₃ is heated?

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Known:

• Mass of calcium carbonate = 24.8 g CaCO₃
• 1 mol CaCO₃ = 100.1 g CaCO₃ (molar mass)
• 1 mol CaO = 56.1 g CaO (molar mass)
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Unknown:

• Theoretical yield of CaO = ? g CaO

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Practice:

1. When 84.8 g of iron (III) oxide reacts with an excess of carbon monoxide, iron is produced.

Fe₂O₃ (s) + 3CO (g) →2Fe(s) + 3CO₂(g)

What is the theoretical yield of iron?

• 59.3 g Fe

Practice:

2. When 5.00 g of copper reacts with excess silver nitrate, silver metal and copper (II) nitrate are produced. What is the theoretical yield of silver in this reaction?

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• 17.0 g Ag

Calculating the Percent Yield of a Reaction

Example

• What is the percent yield if 13.1 g CaO is actually produced when 24.8 g of CaCO₃ is heated?

CaCO₃(s) ^∆ CaO (s) + CO₂ (g)

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Known:

• Actual yield = 13.1 g CaO
• Theoretical yield = 13.9 g CaO (from the previous example)

Unknown:

• Percent yield = ? %

�Calculate:

Practice:

1. If 50.0 g of silicon dioxide is heated with an excess of carbon, 27.9 g of silicon carbide is produced.

• SiO₂ (s) + 3C(s) ^∆ SiC (s) + 2CO (g)

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•  What is the percent yield of this reaction?

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•   83.5%

Practice:

2. If 15.0 g of nitrogen reacts with 15.0 g of hydrogen, 10.5 g of ammonia is produced. What is the percent yield of this reaction?

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• 57.7%

Practice:

3. In a chemical reaction, how does an insufficient quantity of a reactant affect eh amount of product formed?

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• In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms.

Practice:

4. How can you gauge the efficiency of a reaction carried out in the laboratory?

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• The efficiency of a reaction carried out in a laboratory can be measured by calculating the percent yield.

Practice:

5. What is the percent yield if 4.65 g of copper is produced when 1.87 g of aluminum reacts with an excess of copper (II) sulfate?

2Al (s) + 3CuSO₄ (aq) → Al₂(SO₄)₃ (aq) + 3Cu (s)

• 70.5%

Net Ionic Equation

Most reactions happen in a solution of some kind (typically water)

• the way we typically write chemical equations doesn’t fully justify this concept

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• AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

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• Each of these aqueous substances should exist as ions in solution

Net Ionic Equation

In order to describe just the reaction producing the substance of purpose, we can use a Net Ionic Equation, but first we need to write a complete ionic equation.

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• A complete ionic equation describes the entire reaction, separating ions in solution as separate substances.

Complete Ionic Equation

In order to write a complete ionic equation, write each ion present in the solution.

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Ag⁺(aq) + NO₃^-(aq) + Na⁺(aq) + Cl^-(aq) → AgCl(s) + Na⁺ (aq) + NO₃^-(aq)

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* Notice that the AgCl(s) does not separate into ions, this is because it turns into a solid and falls out of solution

Spectator ions

Notice from the equation:

Ag⁺(aq) + NO₃^-(aq) + Na⁺(aq) + Cl^-(aq) → AgCl(s) + Na⁺ (aq) + NO₃^-(aq)

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That the sodium and nitrate ions are present on both sides of the equation.

• An ion that appears on both sides of an equation and is not directly involved in the reaction is called a spectator ion

Net Ionic Equation

In order to change a complete ionic equation into a net ionic equation, we simply remove the spectator ions.

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This leaves the Net Ionic Equation:

Ag⁺(aq) + Cl^-(aq) → AgCl(s)

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*Be sure to balance the equation!!!!!!!!

• Both in charge and number of ions

Net Ionic Equation

A net ionic equation shows only those particles involved in the reaction and is balanced with respect to both mass and charge.

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Practice

Write the balanced net ionic equation for this reaction.

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Ca⁺²(aq) + OH^-(aq) + H⁺(aq) + PO₄^3-(aq) → Ca⁺²(aq) + PO₄^3-(aq) + H₂O(l)

Practice

Write the complete ionic equation and the net ionic equation for the reaction of aqueous calcium hydroxide with phosphoric acid.