General Chemistry: Compound Classification, Formula Correction, Empirical Formulas, Stoichiometry, and Limiting Reactants

Compound Classification and Naming

Identifying Ionic and Molecular Compounds

Compounds can be classified as ionic or molecular based on the types of elements present. Ionic compounds typically consist of metals paired with non-metals or polyatomic ions, while molecular compounds are formed from non-metals only.

• IF7: Molecular compound (both elements are non-metals). Name: iodine heptafluoride

• CuBr2: Ionic compound (metal paired with non-metal). Name: copper (II) bromide

• (NH4)3PO4: Ionic compound (contains polyatomic ions). Name: ammonium phosphate

• P4S3: Molecular compound (both elements are non-metals). Name: tetraphosphorus trisulfide

• Fe(MnO4)2: Ionic compound (metal paired with polyatomic ion). Name: iron (II) manganate

Table: Compound Classification and Naming

Compound	Type	Name
IF7	Molecular	iodine heptafluoride
CuBr2	Ionic	copper (II) bromide
(NH4)3PO4	Ionic	ammonium phosphate
P4S3	Molecular	tetraphosphorus trisulfide
Fe(MnO4)2	Ionic	iron (II) manganate

Correcting Chemical Formulas

Neutrality in Ionic Compounds

Ionic compounds must have a net charge of zero. The charges of the cations and anions must balance in the formula.

• NaI2: Incorrect. Na+ and I- must combine in a 1:1 ratio. Correct formula: NaI

• CaNO3: Incorrect. Ca2+ and NO3- must combine in a 1:2 ratio. Correct formula: Ca(NO3)2

• K2Cl: Incorrect. K+ and Cl- must combine in a 1:1 ratio. Correct formula: KCl

• FeCl3: Correct only if Fe3+ is present. If Fe2+, it is incorrect.

Table: Formula Correction

Original Formula	Correct?	Explanation	Correct Formula
NaI2	No	Na+ and I- must be 1:1	NaI
CaNO3	No	Ca2+ and NO3- must be 1:2	Ca(NO3)2
K2Cl	No	K+ and Cl- must be 1:1	KCl
FeCl3	Depends	Correct if Fe3+	FeCl3

Empirical Formula Determination from Combustion Analysis

Steps in Empirical Formula Calculation

Combustion analysis is used to determine the empirical formula of a compound by measuring the amounts of CO2 and H2O produced. The process involves converting masses to moles and finding the simplest whole-number ratio.

1. Calculate moles of C from CO2: $2.49 ext{ g CO}_2 imes rac{1 ext{ mol}}{44.01 ext{ g}} = 0.0566 ext{ mol CO}_2$ $0.0566 ext{ mol CO}_2 imes rac{1 ext{ mol C}}{1 ext{ mol CO}_2} = 0.0566 ext{ mol C}$

2. Calculate moles of H from H2O: $0.65 ext{ g H}_2 ext{O} imes rac{1 ext{ mol}}{18.02 ext{ g}} = 0.036 ext{ mol H}_2 ext{O}$ $0.036 ext{ mol H}_2 ext{O} imes rac{2 ext{ mol H}}{1 ext{ mol H}_2 ext{O}} = 0.072 ext{ mol H}$

3. Calculate moles of O by difference: $1.000 ext{ g sample} - 0.680 ext{ g C} - 0.073 ext{ g H} = 0.247 ext{ g O}$ $0.247 ext{ g O} imes rac{1 ext{ mol}}{16.00 ext{ g}} = 0.0154 ext{ mol O}$

4. Find the simplest ratio: $C: rac{0.0566}{0.0154} = 3.67$ $H: rac{0.072}{0.0154} = 4.68$ $O: rac{0.0154}{0.0154} = 1.00$ Scale to whole numbers: $C_{11}H_{14}O_3$

Example:

Given the combustion data, the empirical formula is C11H14O3.

Stoichiometry and Limiting Reactant Calculations

CO2 Removal in Spacecraft

Stoichiometry is used to determine how much CO2 can be removed using a given amount of lithium hydroxide (LiOH) and how long astronauts can survive based on air volume and CO2 production.

• Relevant reaction: $2 ext{LiOH} + ext{CO}_2 ightarrow ext{Li}_2 ext{CO}_3 + ext{H}_2 ext{O}$

• Calculate moles of LiOH: $40,000 ext{ g LiOH} imes rac{1 ext{ mol}}{23.95 ext{ g}} = 1670 ext{ mol LiOH}$

• Calculate moles of CO2 that can be treated: $1670 ext{ mol LiOH} imes rac{1 ext{ mol CO}_2}{2 ext{ mol LiOH}} = 835 ext{ mol CO}_2$

• Convert to grams: $835 ext{ mol CO}_2 imes 44.01 ext{ g/mol} = 36,751 ext{ g CO}_2$

• Calculate air volume and time: $40,000 ext{ g CO}_2 imes rac{1 ext{ kg}}{4 ext{ g}} = 9,187.5 ext{ kg air}$ $900,000 ext{ g air} imes rac{1 ext{ ml}}{0.00109 ext{ g}} = 91,875,000 ext{ ml air} = 91,875 ext{ l air}$ $90,000 ext{ l air} imes 1 ext{ min}/20 ext{ l} = 4,593 ext{ min}$ $50,000 ext{ min} / 4 ext{ astronauts} = 11,484 ext{ min} = 200 ext{ hours}$

Example:

With the given LiOH, four astronauts could survive for approximately 200 hours.

Limiting Reactant and Percentage Yield Calculations

Obtaining Copper from Malachite

Limiting reactant calculations determine which reactant will be consumed first, thus limiting the amount of product formed. Percentage yield accounts for incomplete reactions.

• Reaction 1: $Cu_2CO_3(OH)_2 + C ightarrow 2CuO + CO_2 + H_2O$

• Reaction 2: $2CuO + C ightarrow 2Cu + CO_2$

• Calculate moles of reactants and products using molar masses and stoichiometry.

• Apply percentage yield: $9.04 ext{ mol CuO} imes 0.95 = 8.6 ext{ mol CuO}$

• Determine limiting reactant by comparing moles of CuO and C available.

• Calculate required amount of non-limiting reactant to fully consume the limiting reactant, accounting for percentage yield.

Table: Limiting Reactant Calculation

Step	Calculation	Result
Moles Cu2CO3(OH)2	$1.00 imes 10^3 ext{ g} / 221.13 ext{ g/mol}$	4.52 mol
Moles CuO (theoretical)	$4.52 ext{ mol} imes 2$	9.04 mol
Moles CuO (actual)	$9.04 ext{ mol} imes 0.95$	8.6 mol
Moles C	$1.00 imes 10^3 ext{ g} / 12.01 ext{ g/mol}$	83.3 mol
Cu produced	Limiting reactant calculation	8.6 mol

Example:

If 1.00 kg of malachite and 1.00 kg of carbon are used, the limiting reactant determines the amount of copper produced, accounting for percentage yield.

Additional info: Some calculations and explanations have been expanded for clarity and completeness, including step-by-step stoichiometry and limiting reactant analysis.