chem final exam review pt 1 (density -> theoretical yield)

how to find density?

m/v

how to find volume?

m/d

how to find mass?

d*v

what are the steps of giving the formula for a chemical name?

→ tin II chloride)

A) write the charges of each element

B) cross them to turn them into subscripts

→ A) Sn²⁺Cl^-1

B) Sn₁Cl2 → SnCl₂

what are the steps to finding how many atoms in a mole sample of a compound?

→ atoms in a 6.50 mole sample of C₁₂H₂₂O₁₁

A) use the sample of moles given to you

B) multiply the sample by avogadro’s number to get the number of atoms in the compound

→ A) moles * avogadro’s number = atoms

B) 6.50(6.022 × 10²³) = 3.91 × 10²⁴

what are the steps to finding how many atoms of an element are in a sample of moles in a compound?

→ how many O atoms are in 3 moles Al₂O₃?

A) multiply the number of moles given by avogadro’s number

B) multiply by the atoms of the element

→ A) 3 moles (6.022 × 10²³) = 1.80 × 10²⁴

B) (1.80 × 10²⁴) 3 O atoms - 5.42 × 10²⁴

what are the steps to finding how many total atoms are in moles of a compound?

→ how many total atoms in 2.5 moles of Ca₃(PO₄)₂?

A) add up all of the atoms in the compound together

B) multiply the number of moles given by avogadro’s number

C) multiply by the total number of atoms in the compound

→ A) [Ca atoms = 3,Patoms = 2, O atoms = 8] = 13 atoms total

B) 2.5 moles (6.022 × 10²³) = 1.5505 × 10²⁴

C) (1.50 × 10²⁴) 13 = 1.95 × 10²⁴

% composition

% composition = (mass of an element/total molar mass of compound)100%

when given % composition by mass of hydrocarbons, how do you find their empirical formula?

→ 74.83% C, 25.17% H by mass. what is the empirical formula?

A) convert grams to moles for EACH element (% grams/molar mass of element)

B) divide by the smallest number of moles to get a ratio

C) plug into chemical formula

→ A) C: (74.83 g/12.01g/m) = 6.23 moles C, H: (25.17 g/1.00792g/m) = 24.97

moles H

B) C: 6.23/6.23 = 1, H: 24.9/6.23 = 3.99 ~4

C) CH → C₁H₄ → CH₄

how would you find the molecular formula of a compound given the empirical formula and the molecular weight in amu?

→ empirical formula = NO₂, molecular weight = 184.04 amu

A) multiply each amount of atoms in the element by their molar mass to get the molar mass of the entire compound

B) (molecular weight/empirical mass)

C) multiply the outcome to the subscripts in the empirical formula

→ A) N → 1 × 14.01 = 14.01, O → 2 × 16.00 = 32.00; NO₂ = 46.01g/mol

B) (molecular weight/empirical mass) → (184.04amu/46.01g/mol) = 4

C) NO₂(4) → N₄O₈

for balancing coefficients in chemical equations, what are the most important steps?

1) ALWAYS start with the more complex compound, usually polyatomics

2) balance 1 element at a time

3) balance H and O

given a balanced chemical formula, the amount of grams reacted, and molar masses, what are the steps for finding the theoretical yield of a product?

→ 4Fe + 3O₂ → 2Fe₂O₃

115g of Fe & 115g of O₂ reacted

Fe = 55.85g/mol, O₂ = 32.00g/mol, Fe₂O₃ = 159.69g/mol

what is the theoretical yield, in grams, of Fe2O3?

A) convert grams to moles for EACH reactant

B) determine the limiting reagent (whatever you have the least of)

C) using the limiting reagent, calculate the moles of the limiting compound

D) convert moles to grams

→ A) Fe: 115g/55.85g/mol = 2.05 mol Fe, O₂: 115g/32.00g/mol = 3.59 mol O₂

B) 4Fe + 3O₂ means 4 Fe:3O₂ → for every 4 Fe, 3 O₂ is reacted

Fe: 2.05 moles(3O₂/4Fe) → 2.06 moles(3/4) = 1.55 moles O₂ needed

O₂: 3.59 moles(4Fe/3O₂) → 3.59(4/3) = 4.79 moles Fe needed

we have 2.05 moles of Fe and 3.59 moles of O₂. Fe is the limiting reagent because we have the least of it

C) 2.05 moles Fe(2Fe2O3/4Fe) → 2.06(2/4) = 1.03 moles Fe2O3 needed

D) 1.03moles * 159.69g/mol = 164 g Fe₂O₃

theoretical yield = 164 g Fe₂O₃