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Atomic Mass Atoms are so small, it is difficult to discuss how much they weigh in grams. Use atomic mass units. an atomic mass unit (amu) is one twelfth.

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Presentation on theme: "Atomic Mass Atoms are so small, it is difficult to discuss how much they weigh in grams. Use atomic mass units. an atomic mass unit (amu) is one twelfth."— Presentation transcript:

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

4 They are not whole numbers Because they are based on averages of atoms and of isotopes. Carbon =98.89% 12 C 1.11% 13 C <0.01% 14 C Carbon atomic mass = amu

5 The Mole The mole is a number. A very large number, but still, just a number x of anything is a mole A large dozen. The number of carbon atoms in exactly 12 grams of carbon C

6 Molar mass Mass of 1 mole of a substance. Often called molecular weight. To determine the molar mass of an element, look on the table. 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 CH 4 Mg 3 P 2 Ca(NO 3 ) 2 Al 2 (Cr 2 O 7 ) 3 CaSO 4 2H 2 O

8 Practice Vitamin C (C 9 H 8 O 6 ) is an essential vitamin. What is its molar mass? If a tablet contains mg of Vitamin C, how many moles and how many molecules of Vitamin C does it contain? #45

9 Practice The molecular formula for Aspartame is C 14 H 18 N 2 O 5 A) Calculate its molar mass. B) How many moles are in 10.0 g? C) What is the mass of 1.0 x 10 9 molecules? D) How many atoms of nitrogen are in 1.2 g? E) What is the mass of one molecule of aspartame? #51

10 Percent Composition Percent of each element a compound is composed of. Find the mass of each element, divide by the total mass, multiply by a 100. Easiest if you use a mole of the compound. Find the percent composition of CH 4 Al 2 (Cr 2 O 7 ) 3 CaSO 4 · 2H 2 O

11 Empirical Formula Determination Base calculation on 100 grams of compound. Determine moles of each element in 100 grams of compound. Divide each value of moles by the smallest of the values. Multiply each number by an integer to obtain all whole numbers. A sample is 59.53% C, 5.38%H, 10.68%N, and 24.40%O. What is its empirical formula.

12 Empirical Formula Determination Percent to mass Mass to mole Divide by small Times till whole.

13 Problem A gram sample of a compound (vitamin C) composed of only C, H, and O is burned completely with excess O g of CO 2 and g of H 2 O are produced. What is the empirical formula?

14 Empirical To Molecular Formulas Empirical is lowest ratio. Molecular is actual molecule. Need Molar mass. Ratio of empirical to molar mass will tell you the molecular formula. Must be a whole number.

15 Problem Determine the molecular formula from the empirical formula and it molar mass: SNH ( g/mol), NPCl 2 ( g/mol) #62 A compound is made of only oxygen and nitrogen. It is 30.4% N by mass. Its molar mass is 92 g/mol. What is its empirical and molecular formula? #67

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

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

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

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

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

21 Abbreviations (s)  (g)  (aq) heat  catalyst

22 Practice Ca(OH) 2 + H 3 PO 4  H 2 O + Ca 3 (PO 4 ) 2 Cr + S 8  Cr 2 S 3 KClO 3 (s)  KCl(s) + O 2 (g) Fe 2 O 3 (s) + Al(s)  Fe(s) + Al 2 O 3 (s) Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas. Aqueous barium hydroxide reacts with aqueous sulfuric acid to produce barium sulfate and water. #75

23 Meaning A balanced equation can be used to describe a reaction in molecules and atoms. C 2 H 5 OH + 3O 2  2CO 2 + 3H 2 O The equation is balanced. 1 mole of ethanol reacts with 3 moles of oxygen. --> to produce 2 moles of carbon dioxide and 3 moles of water.

24 Stoichiometry 1.Balance the equation. 2.Convert mass to moles. 3.Set up mole ratios. 4.Use mole ratios to calculate moles of desired substituent. 5.Convert moles to grams, if necessary.

25 Examples One way of producing O 2 (g) involves the decomposition of potassium chlorate into potassium chloride and oxygen gas. A 25.5 g sample of Potassium chlorate is decomposed. How many moles of O 2 (g) are produced? How many grams of potassium chloride? How many grams of oxygen?

26 Examples K 2 PtCl 4 (aq) + NH 3 (aq)  Pt(NH 3 ) 2 Cl 2 (s) + KCl(aq) #91 What mass of Pt(NH 3 ) 2 Cl 2 can be produced from g of K 2 PtCl 4 ? How much KCl will be produced? How much of Pt(NH 3 ) 2 Cl 2 would form from 65.0 grams of NH 3 ?

27 Limiting Reagent Reactant that determines the amount of product formed. The one you run out of first. Makes the least product. Book shows you a ratio method. It works, but you are used to finding grams.

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

29 Example Ammonia is produced by the following reaction N 2 + H 2  NH 3 What mass of ammonia can be produced from a mixture of 100.g N 2 and 5.00g H 2 ? How much unreacted material remains?

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

31 Percent Yield % yield = Actual x 100% Theoretical % yield = what you got x 100 what you should have got

32 Examples Aluminum burns in bromine producing aluminum bromide. In a laboratory 6.0 g of aluminum reacts with excess bromine g of aluminum bromide are produced. What are the three types of yield?

33 Examples Years of experience have proven that the percent yield for the following reaction is 74.3% Hg + Br 2  HgBr 2 If 10.0 g of Hg and 9.00 g of Br 2 are reacted, how much HgBr 2 will be produced? If the reaction did go to completion, how much excess reagent would be left?

34 Examples Commercial brass is an alloy of Cu and Zn. It reacts with HCl by the following reaction. Zn(s) + 2HCl(aq)  ZnCl 2 (aq) + H 2 (g) Cu does not react. When g of brass is reacted with excess HCl, g of ZnCl 2 are eventually isolated. What is the composition of the brass?


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