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STOICHIOMETRY Study of the amount of substances consumed and produced in a chemical reaction.

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**The Mole and Stoichiometry**

a. SI unit for a given amount of substance b. abbreviation – mol c. standard mole = the amount of a substance that contains as many particles as there are found in exactly 12 g of carbon-12

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**The Mole and Stoichiometry**

2. Avogadro’s number – a. the number of particles in exactly one mole of a pure substance b x 1023 c. 12 g Carbon-12 = one mol = x atoms of Carbon-12

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**The Mole and Stoichiometry**

3. Molar mass – a. mass of one mole of a pure substance b. unit g/mol c. for elements found on the periodic table d. Examples: nitrogen: g/mol (1 mol of nitrogen = 14.0 grams = 6.02 x 1023 atoms nitrogen) silver: g/mol (1 mol of silver= grams = 6.02 x 1023 atoms silver)

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**The Mole and Stoichiometry**

e. Calculating Molar masses for compounds 1. Identify the name and number of each atom found in the compound 2. multiply the number of atoms by atomic mass for that element (found on the Periodic table). 3. add all the masses together

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**The Mole and Stoichiometry**

f. Example: H2O : Al2 (PO4) 3

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**The Mole and Stoichiometry**

4. Gram/Mole Conversions a. Mass to moles Given: grams / molar mass = moles Example: How many moles are there in 12 g of MgS?

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**The Mole and Stoichiometry**

b. Moles to mass Given moles X molar mass = grams Example: What is the mass of 2.5 moles of FeCl3?

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**The Mole and Stoichiometry**

5. Mass/Mass Conversions (Grams A)( 1 mol A) (coefficient B)( mass B) ( )( mass A )(Coefficient A)(1 mol B) Example 2H2 + O2 2H2O How many grams of water is produced from 5.00 g hydrogen?

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**STOICHIOMETRY Setting the problem up: (mass-mass)**

1. Write and balance the equation.. 2. Identify the given and unknown. 3. Identify the mole ratio (coefficients from the balanced equation for the given and unknown). 4. Calculate the molar mass for the given and unknown.

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SOLVING THE PROBLEM: 1. Convert grams of given to moles of given (Divide by molar mass of given) 2. Convert moles of given to moles of unknown (Multiply by the mole ratio- coefficient of unknown/coefficient of given) 3. Convert moles of unknown to grams of unknown(Multiply by the molar mass of unknown)

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Limiting Reactants A. Limiting Reactant – the reactant that limits the amount of the reactant that can combine, and the amount of product formed in a chemical reaction B. Excess Reactant – the substance that is not used up completely in a reaction

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Limiting Reactants C Problem-solving steps for determining limiting/excess reactant 1. Write and balance the equation 2. Identify the amounts of the two reactants and unknown 3. Identify mole ratio 4. Calculate the molar mass for the reactants and unknown

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Limiting Reactants 5. Determine moles of both reactants by dividing mass (given in problem) by molar mass (step 4) 6. Choose one of the reactants as reactant A and the other as reactant B 7. Multiply the moles of reactant A by the mole ratio of reactant B/reactant A.

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Limiting Reactants 8. Compare the moles of reactant B needed (answer from step 7) to actual number of moles present (answer from step 5). If amount is less than what you have then reactant B is in excess and reactant A is the limiting reactant If amount needed is more than what you have then reactant B is the limiting reactant and reactant A is in excess.

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**Determining the amount produced**

1. Find the limiting reactant 2. __________ the moles of reactant (calculated in Step C5 above) by the mole ratio of product/reactant 3. Multiply moles of product (answer from step D2) by the molar mass of the _________

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IV. Percent Yield 1. Theoretical yield – the amount of product predicted to be produced on the basis of the balanced chemical equation 2. Actual yield – the amount of product actually obtained in a reaction actual yield % yield = theoretical yield X 100

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**Percent Composition Steps in finding Percent Composition:**

#1 Calculate the molar mass of the compound #2 Divide the mass of the element by the Total #3 Multiply by 100. #4 Check answer: the Parts should add up to 100.

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Percent Composition Example: Find the percent composition of each element in carbon dioxide.

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Empirical Formulas Empirical Formula – Lowest whole number formula for a compound

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**Empirical Formulas Steps in solving:**

#1 Convert grams of given to moles (divide by atomic mass) # 2 Find mole ratio – divide all parts by the smallest # 3 IF the ratio is a whole number – then place these in the formula as subscripts HOWEVER, if you have fractions- MUST multiply all by a common factor to arrive at a small whole number

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Empirical Formulas EXAMPLE 1: A certain compound contains 52% nickel, 9.6% carbon and 38.4%oxygen. What is its empirical formula?

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Empirical Formulas EXAMPLE 2: A compound is found to contain 43.7% phosphorus and 56.9% oxygen. Calculate its simplest formula.

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Molecular Formulas Molecular formula – whole number multiples of an empirical formula: The molecular formula is the true formula Example: HO (empirical) H2O2 (Molecular)

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**Molecular Formulas Steps in solving:**

#1 Molecular mass = x Empirical mass # 2 (Empirical formula) x

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Molecular Formulas Example : The molecular mass for a hydrocarbon was found to be 64 g/mol. If the empirical formula is CH4, what is the molecular formula?

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CHAPTER 3b Stoichiometry.

CHAPTER 3b Stoichiometry.

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