ch 3 mass relations in chemistry stoichiometry n.
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CH 3 Mass Relations in Chemistry; Stoichiometry. Atomic Mass. Indicates how heavy an element is compared to another element. Units AMU---Atomic Mass Unit Defined as 1/12 of the mass of a C-12 atom. Atomic Mass from isotope composition. Isotopic Abundance: the natural of an isotope.

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atomic mass
Atomic Mass
  • Indicates how heavy an element is compared to another element.
  • Units AMU---Atomic Mass Unit
    • Defined as 1/12 of the mass of a C-12 atom
atomic mass from isotope composition
Atomic Mass from isotope composition
  • Isotopic Abundance: the natural of an isotope.
mass of individual atoms
Mass of individual atoms
  • Mass of 1 atom = molar mass/ NA

(Avogadro's #)

Reacquaint yourself with the mole wheel.

the mole
The Mole
  • 1 mol= 6.022 x 1023 items
    • 1mol H = 6.022 x 1023 H atoms = 1.008g
    • 1mol Cl= 6.022 x 1023 Cl atoms = 35.45g
    • 1mol Cl2= 6.022 x 1023 Cl2 molecules = 70.90g
molar mass mm
Molar mass (MM)
  • Molar mass is numerically equal to the sum of the atomic masses.
mass from formula
Mass % from formula
  • % composition of K2CrO4

Use part / whole, assume you have 1 mole of compound. (the math is easier)

1mol = K2CrO4194.20 g

%K = 78.90/194.20 =

%Cr =

%O =

simplest empirical formula
Simplest / empirical formula
  • Simplest whole number ratio of atoms present in a compound.
simplest empirical formula from composition
Simplest (empirical) formulafrom % composition
  • Steps:
    • Find the mass of each element in the sample compound, assume 100g total.
    • Find the numbers of moles of each compound.
    • Divide each by the smallest # of moles and look for obvious ratios.

Use the following K= 26.6%, Cr= 35.4%, O = 38.0%

simplest formula empirical from analytical data
Simplest formula (empirical) from analytical data

An organic sample containing only C, H, O atoms weighs 1.000g

Burning the sample gives 1.466g CO2,

0.6001g H2O

Find the simplest formula…

  • All the carbon from the sample is “locked up” in CO2
  • The portion of CO2 that is carbon can be determined by the mass ratio in the formula (12.01/44.01)
  • This multiply this by the mass of CO2 and you find grams of carbon in the sample.
yield of product in a reaction
Yield of product in a reaction
  • Ordinarily, reactants are not present in the exact ratio required for reaction.

Usually 1 is in excess; some left when reaction is over.

1 is limiting; completely consumed to give the theoretical yield of product.

calculating theoretical yield
Calculating theoretical yield
  • Calculate the yield expected if the first reactant is limiting.
  • Repeat the calculation for the second reactant
  • The theoretical yield is the SMALLER of these two quantities. The reactant that gave the smaller theoretical yield is the limiting reactant.
2 ag s i 2 s 2 agi s
2 Ag (s) + I2 (s)  2 AgI (s)
  • Calculate the theoretical yield of AgI and determine the limiting reactant.
      • There is 1.00g Ag, and 1.00g I2.
yield
% Yield
  • Suppose the actual yield is 1.50g AgI, what was the % yield?

Actual/Theoretical (x100) = % Yield