Concepts to Know for Exam 1
Introduction, Stoichiometry, Empirical Formula
Stoichiometry: quantitative relationship b/w elements in compound or b/w
compounds/elements in chemical reactions
Quantitative Analysis: determination of the amount by weight of each element or compound
present
Significant Figures:
Multiplying/Dividing:
● Answer must have no more sig figs than either # multiplied
● Must be lowest amount b/w 2 multiplied #’s
Addition/Subtration:
● Only as precise as the least precise thing you add/subtract
Atoms → Grams
● Use Avogadro’s #: 6.022e23 atoms
● Multiple atoms by Avogadro’s number
Elemental Composition (Mass %)
% A = (mass A/total mass) x 100
Empirical Formula: the simplest formula, only tells you the ratio of elements in a compound
● Find the moles of each element in formula, then divide by smallest #
● When you find the moles, you get the mole ratio of each element in the compound
● Make sure the empirical formula ends in whole numbers!
Molecular Formula: the actual # of atoms in each element in a compound
● Need molecular weight to know molecular formula
Chemical Formula: the actual # of atoms of each element in a compound
Equations for Chemical Reactions
Balancing Chemical Equations
General Process (non-mathematically):
● Count up number of atoms in the chemical reaction
● Add one coefficient at a time, starting with most complex compound
● Check when finished! Ensure atoms in = atoms out; charge in = charge out
General Process (mathematically)
● Instead of writing a number coefficient, write a letter (a, b, c, d)
● Write an equation for each element to calculate coefficients
, Notes:
Don’t alter the subscripts, only alter the coefficients preceding the atoms or molecules
Calculating Mass of Reactant
EX: How many grams of SiO2 are required to synthesize 1.5g of MgSiO3?
10 C + 6 SIO2 + 2 Mg3(PO4)2 → P4 + 6 MgSiO3 + 10 CO
1.5 g MgSiO3 x 1 mol MgSiO3/100.4 g MgSiO3 x 6 mols SiO2/6 mols MgSiO3 x 60.1 g SiO2/1
mol SiO2 = 0.90g SiO2
Molar mass of MgSiO3
Stoichiometric Coefficient (gained from looking at equation)
In Simpler Terms:
grams A ÷ MW A = mol A x mol B/mol A = mol B x MW B = grams B
*MW is molecular weight
Limiting Reagent: determines the amount of products that will be produced
● To identify, find the compound that gives the LEAST amount of product
Limiting Reagent & Reaction Table:
EX: You have 15g of P4O6 and 20g of H2O. How many grams of H3PO3 can you synthesize?
P4O6 + 6 H2O → 4 H3PO3
1) Balance equation
2) Convert starting amounts to moles
3) Determine limiting reagent
4) Make Reaction Table (starting with limiting reagent; the delta change of this one will = 0
final)
5) Use stoich coefficients from balanced equation to determine delta change of other
reactants and products
6) Convert final moles back to masses (grams)
Theoretical Yield: amount of product based on the limiting reactant; the maximum that could be
made
Actual Yield: the amount of product that is actually made
Percent Yield: actual yield/theoretical yield x 100
*Calculate % yield from grams or moles
Percent Excess: amount leftover/amount used up x 100
*Calculate % excess from grams or moles
Introduction, Stoichiometry, Empirical Formula
Stoichiometry: quantitative relationship b/w elements in compound or b/w
compounds/elements in chemical reactions
Quantitative Analysis: determination of the amount by weight of each element or compound
present
Significant Figures:
Multiplying/Dividing:
● Answer must have no more sig figs than either # multiplied
● Must be lowest amount b/w 2 multiplied #’s
Addition/Subtration:
● Only as precise as the least precise thing you add/subtract
Atoms → Grams
● Use Avogadro’s #: 6.022e23 atoms
● Multiple atoms by Avogadro’s number
Elemental Composition (Mass %)
% A = (mass A/total mass) x 100
Empirical Formula: the simplest formula, only tells you the ratio of elements in a compound
● Find the moles of each element in formula, then divide by smallest #
● When you find the moles, you get the mole ratio of each element in the compound
● Make sure the empirical formula ends in whole numbers!
Molecular Formula: the actual # of atoms in each element in a compound
● Need molecular weight to know molecular formula
Chemical Formula: the actual # of atoms of each element in a compound
Equations for Chemical Reactions
Balancing Chemical Equations
General Process (non-mathematically):
● Count up number of atoms in the chemical reaction
● Add one coefficient at a time, starting with most complex compound
● Check when finished! Ensure atoms in = atoms out; charge in = charge out
General Process (mathematically)
● Instead of writing a number coefficient, write a letter (a, b, c, d)
● Write an equation for each element to calculate coefficients
, Notes:
Don’t alter the subscripts, only alter the coefficients preceding the atoms or molecules
Calculating Mass of Reactant
EX: How many grams of SiO2 are required to synthesize 1.5g of MgSiO3?
10 C + 6 SIO2 + 2 Mg3(PO4)2 → P4 + 6 MgSiO3 + 10 CO
1.5 g MgSiO3 x 1 mol MgSiO3/100.4 g MgSiO3 x 6 mols SiO2/6 mols MgSiO3 x 60.1 g SiO2/1
mol SiO2 = 0.90g SiO2
Molar mass of MgSiO3
Stoichiometric Coefficient (gained from looking at equation)
In Simpler Terms:
grams A ÷ MW A = mol A x mol B/mol A = mol B x MW B = grams B
*MW is molecular weight
Limiting Reagent: determines the amount of products that will be produced
● To identify, find the compound that gives the LEAST amount of product
Limiting Reagent & Reaction Table:
EX: You have 15g of P4O6 and 20g of H2O. How many grams of H3PO3 can you synthesize?
P4O6 + 6 H2O → 4 H3PO3
1) Balance equation
2) Convert starting amounts to moles
3) Determine limiting reagent
4) Make Reaction Table (starting with limiting reagent; the delta change of this one will = 0
final)
5) Use stoich coefficients from balanced equation to determine delta change of other
reactants and products
6) Convert final moles back to masses (grams)
Theoretical Yield: amount of product based on the limiting reactant; the maximum that could be
made
Actual Yield: the amount of product that is actually made
Percent Yield: actual yield/theoretical yield x 100
*Calculate % yield from grams or moles
Percent Excess: amount leftover/amount used up x 100
*Calculate % excess from grams or moles
