Calculations with chemical formulas and equations

Question 1

Molecular Mass

Answer 1

• Sum of the atomic masses of all the atoms in a molecule of the substance.
• Molecular mass of ethylene, C₂H₄
  
  • Carbon: 2 • 12.01 amu = 24.02 amu
  • Hydrogen: 4 • 1.008 amu = 4.032 amu
  • Total: 28.05 amu
• Mass spectrometer is used to determine molecular mass.

Question 2

Formula mass

Answer 2

• The sum of all the atomic masses of all atoms or ions present in a formula unit of any compound, whether molecular or not.

Formula mass of Na₂SO₄

• Sodium: 2 • 22.99 amu = 45.98 amu
• Sulfur: 1 • 32.07 amu = 32.07 amu
• Oxygen: 4 • 16.00 amu = 64.00 amu
• Total: 142.1 amu

Question 3

Mole

Answer 3

the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12g of carbon-12

Question 4

Avogadro’s number

Answer 4

• The number of atoms in a 12 g sample of carbon-12
• Named after the Itailian scientist, Amadeo Avagadro (1776-1856)
• 6.0221367 x 10²³ equals to 1 mole.

Question 5

Mole Calculations

Answer 5

• enables to inter-convert from moles to grams, grams to moles and find the number of molecules in a given mass.
• molar mass = mass/moles

Question 6

Percent Composition

Answer 6

• Mass percentage of each element in the compound.
• Proportion of the constituent elements expressed as the number of grams od each element per 100 grams of the compound.
• Knowing the total mass, we can determine the contribution of each element to the total

Question 7

Mass Percentage from the Formula

Answer 7

parts of A per hundred parts of the total, by mass

• Steps
  
  • Find Molar mass
  • Divide total mass of each element in the formula by the molar mass and multiply by 100.

Question 8

Emperical Formula

Answer 8

• Also known as simplest formula
• smallest whole number ration of atoms present in a compound.
• Sometimes the empirical formula and the molecular formula is the same, but most of the time, the molecular formula is a whole number multiple of the empirical formula.

Question 9

Empirical Formula from the composition

Answer 9

• Either from the masses of elements or from the percent composition.
• steps
  
  • assume 100.0 g of the substance
  • convert the gram of each element into moles using each element’s molar mass
  • divide each value by the smallest value
    
    • if whole numbers, use them as subscripts
    • if not whole numbers, multiply by the smallest number that would convert them to whole numbers.

Question 10

What is the simple rhyme to remember to determine empirical formula.

Answer 10

• Percent to mass
• mass to mole
• divide by small
• multiply until whole.

Question 11

Molecular formulas from empirical formula

Answer 11

• Molecular mass = n x (empirical formula mass)

Question 12

Stoichiometry

Answer 12

• Calculation of the quantities of reactants and products involved in a chemical reaction.
• Quantitative relationship among reactants and products

Question 13

Molar interpretation of chemical equation

Answer 13

• A chemical equation can be interpreted in terms of molecules, ions, formula units and moles.
• 2H₂ + O₂ → 2H₂O can be interpreted as
  
  • 2 moles of H₂ + 1 mole of O₂ → 2 moles of H₂O
  • 2 grams of H₂ + 1 gram of O₂ → 2 grams of H₂O

Question 14

Mole ratio

Answer 14

the ratio between the number of moles of any two species in a chemical reaction.

Question 15

Limiting Reactant

Answer 15

• Also known as Limiting reagent
• Limits the amount of product that can be formed
• It is completly used up when a reaction is complete
• The moles of products are always determined by the starting moles of the limiting reagent.
• To determine the limiting reagent, use stoichiometry to find the product for each of the element. The smallest amount is the limiting reagent.

Question 16

Theoretical Yield

Answer 16

Maximum amount of product that can be obtained from given amounts of reactants according to the chemical equation.

Question 17

Actual Yield

Answer 17

Measured mass of the product

Note: actual yield may be less than the theoretical yield because of:

• Many reactions do not go to completion
• Two or more reactions may occur simutaneously producing desired and undesired products
• Accurate amount of product may be difficult to measure
• starting materials may not be pure.

Question 18

Percentage Yield

Answer 18

Ratio of actual yield to theoretical yield and multiplied by 100

Both the actual and theoretical yield must have the same units to find percentage