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AP Chemistry West Valley High School Mr. Mata. Atomic Mass l Atoms are so small, it is difficult to discuss how much they weigh in grams. l Use atomic.

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Presentation on theme: "AP Chemistry West Valley High School Mr. Mata. Atomic Mass l Atoms are so small, it is difficult to discuss how much they weigh in grams. l Use atomic."— Presentation transcript:

1 AP Chemistry West Valley High School Mr. Mata

2 Atomic Mass l Atoms are so small, it is difficult to discuss how much they weigh in grams. l Use atomic mass units. l an atomic mass unit (amu) is 1/12 th the mass of a carbon-12 atom. l This gives us a basis for comparison. l The decimal numbers on the Periodic Table are atomic masses in amu.

3 They are not whole numbers l based on averages of atoms & isotopes. l figure average atomic mass from mass of the isotopes & relative abundance. l add up % as decimals x isotope masses.

4 Example l There are two isotopes of carbon 12 C with a mass of amu(98.892%), and 13 C with a mass of amu (1.108%) l 12 C: x = l 13 C : x = l Total atomic mass = amu for C

5 The Mole l The mole is a very large number. l 6.02 x of anything is a mole. l The number of atoms in exactly 12 grams of carbon-12. l Makes the numbers on the table the mass of the average atom.

6 Molar mass l mass of 1 mole of a substance. l often called molecular weight or gram formula mass. l To determine the molar mass of an element, look on the table. l To determine the molar mass of a compound, add up the molar masses of the elements that make it up.

7 Find the molar mass of l CH (4) = g l Mg 3 P (3) (2) = g l Ca(NO 3 ) (2) (6) = g l Al 2 (Cr 2 O 7 ) (2)+52.00(6)+16.00(21) = g l CaSO 4 · 2H 2 O (4) (4) (2) = g

8 Percent Composition l Percent of each element in a compound. l Find the mass of each element, divide by the total mass, multiply by a 100. l Easiest if you use a mole of the compound.

9 Percent Composition l Ex: Find the % composition of CH 4 l Molar mass of = CH 4 = grams l C = grams x 100 = 74.88% grams l H = grams x 100 = 25.12% grams

10 Working backwards l From percent composition, you can determine the empirical formula. l Empirical Formula the lowest ratio of atoms in a molecule. l Based on mole ratios.

11 Empirical Formula l A sample is 71.65% Cl, 24.27%C, and 4.07%H. Find empirical & molecular formulas? (molar mass = g/mol) l Step 1: convert mass % to mass (g). l Cl: g Cl x 1 mol Cl = mol Cl g Cl l C: g C x 1 mol C = mol C g C

12 Empirical Formula l H: 4.07 g H x 1 mol H = 4.04 mol H g H l Step 2: Divide each mole value by the smallest number of moles present.

13 Empirical Formula l mol Cl = 1 mol Cl mol l mol C = 1 mol C mol l 4.04 mol H = 1.99 mol H = 2 mol H mol l Empirical formula = ClCH 2

14 Molecular Formula l Empirical Formula = ClCH 2 l To find molecular formula, compare empirical formula with molar mass: l Molar mass = g/mol = 2 Empirical mass g/mol Molecular formula = (ClCH 2 ) 2 = Cl 2 C 2 H 4

15 Chemical Equations l Are sentences. l Describe what happens in a chemical reaction. Reactants  Products l Equations should be balanced. l Have the same number of each kind of atoms on both sides.

16 Balancing equations CH 4 + O 2  CO 2 + H 2 O ReactantsProducts C11 O23 H42

17 Balancing equations CH 4 + O 2  CO H 2 O ReactantsProducts C11 O23 H424

18 Balancing equations CH 4 + O 2  CO H 2 O ReactantsProducts C11 O23 H424 4

19 Balancing equations CH 4 + 2O 2  CO H 2 O ReactantsProducts C11 O23 H

20 Abbreviations ( s ),  for product) ( g ),  for product) l ( aq ) l heat  l catalyst

21 Meaning l A balanced equation can be used to describe a reaction in molecules and atoms. l Not grams. l Chemical reactions happen molecules at a time. l or dozens of molecules at a time. l or moles of molecules.

22 Stoichiometry l Given an amount of either starting material or product, determining the other quantities. l use conversion factors from: –molar mass (g - mole) –balanced equation (mole - mole)

23 Limiting Reagent l Reactant that determines the amount of product formed. l The one you run out of first. l Makes the least product. l Book shows you a ratio method.

24 Limiting reagent l To determine the limiting reagent requires that you do two stoichiometry problems. l Figure out how much product each reactant makes. l The one that makes the least is the limiting reagent.

25 Excess Reagent l The reactant you don’t run out of. l The amount of stuff you make is the yield. l The theoretical yield is the amount you would make if everything went perfect. l The actual yield is what you make in the lab.

26 Percent Yield l % yield = Actual x 100% Theoretical l % yield = what you got x 100% what you could have got


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