Chemistry 11 Unit 3 Quantities in Chemical Reactions Flashcards Preview

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Flashcards in Chemistry 11 Unit 3 Quantities in Chemical Reactions Deck (10):



Mole: (mol)-used to described an amount of atoms or molecules.
• 1 mole = 6.02 x 10^23
• Referred to Avogadro’s number (NA)

Ex: 12 g of carbon (pencil leads) = 1 mole (atomic mass of an element on the Periodic Table)
-602 earth sized planets covered 1 m deep of peas →large number
-1/8 the surface area of the sun


Important things to remember when doing calculations:

-Balance all equations
-proper rounding (final: 2 decimal; other: 3 or more decimal)
-be aware of what the question is asking for
-be aware of numbers (0.0___ on calculator)


Atomic Mass

Atomic Mass: (g/mol)- mass of one mole of atoms in an element (mass of an atom)
• Contains 6.02 x 1023 atoms
• Ex: Carbon’s atomic mass is 12 g = 1 mole
• Sum of the protons and neutrons in an atom (because make up majority of atom’s mass) →proton and neutron each have a mass of 1
• Since electrons are extremely light, do not contribute to mass


Molar Mass

Molar mass: (g/mol)- mass of one mole of molecules in a compound or substance.
• Contains 6.02 x 1023 molecules.
• Found by adding the atomic masses of atoms making up the molecule/compound
o Ex: water
• 2(H) + 1(O)
• 2(1) + 1(16)
• 18 g/mol
• Mass on the periodic table (ex: 16g is 1 mole for oxygen)
• Do not need to use coefficient (because coefficient is only used for mole ratios)


Mole calculations (n)


n = V/MV (molar volume at STP: 22.4 L/mol)

n = m/mm

n = #p/Na (avogadro's number in particles)

# of atoms
1) find #p (molecules)
2) multiply #p by subscript of element want to find in the compound


Calculation Reminders

-gas molecules = 2 atoms (Ex: Fluorine gas = F2 with molar mass 2(19) = 38 g/mol)
-write down what is given, need to find



-requires balanced chemical reaction
-coefficients in front of substances represent # of moles

Mole ratio: (mole of substance you have/coefficient of substance you have)*coefficient of substance you want

Mole to Mole -->using mole ratio
Mixed mass-volume-particle


Energy Stoichiometry

-exothermic (releasing heat, product side)
-endothermic (absorbing heat, reactant side)

1) convert to moles
2) mole ratio x energy coefficient
3) answer above in KJ

Alternative way to solve:
Coefficient of known mole substance : coefficient KJ

mole of known substance : X (unknown)

^solve for X as a ratio


Limiting Reactants

Reactant that limits the final yield of products (ex: cheese sandwich analogy)
-occurs when given two masses or two moles
-requires extra step before stoichiometry
-Goal is to determine the LR out of the two reactants (one will be LR one will be excess)

1) take info on reactants and convert to moles
2) box 1:1 ratio
for each reactant:
# of moles/coefficient
-lowest/smallest number of moles is LR

4)use the moles of the LR to solve

To calculate amount of excess left over:
Moles of excess to start - moles of excess used/required (from LR moles/LR coefficient * excess coefficient)
-leftover mols and grams (must convert)

Other ways of saying left over: unreacted, not used, amount left over


Percent Yield

Amount of product obtained/produced in a chemical reaction

% yield = actual value (from experiment)/theoretical value (stoichiometry)*100