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 one.

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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 one twelth the mass of a carbon-12 atom. l The decimal numbers on the table are atomic masses in amu.

They are not whole numbers l Because they are based on averages of atoms and of isotopes.

mass 10000 Cu atom 6909 Cu-63 6909x62.93 =434783.37 u 3091 Cu-65 3091x64.93 =200698.63 u Total mass= 635482 u Average mass= total mass/10000=63.55 u

The Mole l The mole is a number. l A very large number, but still, just a number. l 6.022 x 10 23 of anything is a mole l A large dozen. l The number of atoms in exactly 12 grams of carbon-12.

The Mole l 1 mol of atoms of any element weighs the number value of atomic weight in g. l 1 mol Cu weighs 63.55 g l 1 mol Ga weighs 69.723 g

Molar mass, M l Mass of 1 mole of a substance. l Often called molecular weight. 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.

Find the molar mass of CH 4 : CH4 → C + 4 H m CH4 = m C + m H M CH4 = M C + 4×M H Mg 3 P 2 : M = 3 × M Mg + 2 × M P l Ca(NO 3 ) 3 l Al 2 (Cr 2 O 7 ) 3 l CaSO 4 · 2H 2 O

Number of moles, n Avogadro’s number N av =6.02x10 23 mol -1

Number of moles, number of atoms in 10 g Al? Number of moles, mass of 5.00x10 20 atom Co?

Number of molecules of isopentyl acetate in 1 ug? Number of carbon atoms?

Mass of CO 3 2- in 4.86 mol CaCO 3 ?

Percent Composition l Percent of each element a compound is composed of. l What is the percent composition of C 2 H 5 OH?

Suppose we have 1 mol C 2 H 5 OH Sum of all percentages=100% 100 g of C 2 H 5 OH contains 52.14 g C, 13.13 g H and 34.73 g O.

Find the percent compositions of the following compounds: l Al 2 (Cr 2 O 7 ) 3 l CaSO 4 · 2H 2 O

l From percent composition, you can determine the empirical formula. l Empirical Formula the lowest ratio of atoms in a molecule. A sample is 59.53% C, 5.38%H, 10.68%N, and 24.40%O, what is its empirical formula? CHNO sample mass=100 g59.53 g5.38 g10.68 g24.4 g n=m/M59.53/12.015.38/1.00810.68/14.0024.4/16.00 n4.965.340.761.53 Divide by smallest value6.537.0312.01 E.F. C 13 H 14 N 2 O 4

Empirical Formula and Molecular Formula l H 2 O 2 EF: HO l Such compound doesn’t exist! l MF = n × EF H 2 O 2 = 2 (HO) l M MF =n × M EF 34 = 2 × 17 l 2×M H + 2×M O = 2 (M H +M O ) l Alkenes: C 2 H 4, C 3 H 8, C 4 H 8, C 5 H 10 … l EF : CH 2

ClCH sample mass=100 g71.65 g24.27 g4.07 g n=m/M71.65/35.4524.27/12.014.07/1.008 n2.021 4.04 Divide by smallest value112 E.F. CH 2 Cl M.F. C 2 H 4 Cl 2

Elemental Analysis A substance is known to be composed of C, H and nitrogen. When 0.1156 g of this substance is reacted with oxygen, 0.1638 g CO 2 and 0.1676 g H 2 O. Find the empirical formula.

l Assumptions: 1) All C in sample converted into CO 2.

2) All H in sample converted into H 2 O.

CHN mass0.04470 g0.01875 g0.0521 g n=m/M0.04470/12.010.01875/1.0080.0521/14.00 n0.0037220.018600.003721 Divide by smallest value151 E.F. CH 5 N

Chemical Equations l Are sentences that describe what happens in a chemical reaction. Reactants  Products CH 4 + O 2  CO 2 + H 2 O l Redistribution of bonds, atoms don’t change! l Equations should be balanced: Have the same number of each kind of atoms on both sides. l Law of conservation of mass. CH 4 + 2O 2  CO 2 + 2 H 2 O Balancing by inspection.

Abbreviations ( s )  ( g )   l  l ( aq ) l heat  l catalyst

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 )  Cl 2 ( g ) + O 2 ( g ) l Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas. Fe 2 O 3 ( s ) + Al( s )  Fe( s ) + Al 2 O 3 ( s )

Stoichiometric Calculations What mass of oxygen will react completely with 96.1 g of propane? C 3 H 8 + O 2  CO 2 + H 2 O l Balance the equation! C 3 H 8 + 5 O 2  CO 2 + 4 H 2 O l Language of moles, not grams!

C 3 H 8 + 5 O 2  CO 2 + 4 H 2 O 1 5 2.18 ?

Grams of CO 2 produced? C 3 H 8 + 5 O 2  CO 2 + 4 H 2 O 1 3 2.18 ?

Mass of CO 2 absorbed by 1.00 kg LiOH (s) ? 2 1 41.8 ?

Antiacids

1 0.0119 ? 1 2 0.0172 ?

Examples l 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? l How many grams of potassium chloride? l How many grams of oxygen?

Examples l A piece of aluminum foil 5.11 in x 3.23 in x 0.0381 in is dissolved in excess HCl(aq). How many grams of H 2 ( g ) are produced? How many grams of each reactant are needed to produce 15 grams of iron form the following reaction? Fe 2 O 3 ( s ) + Al( s )  Fe( s ) + Al 2 O 3 ( s )

Examples K 2 PtCl 4 ( aq ) + NH 3 ( aq )  Pt(NH 3 ) 2 Cl 2 ( s )+ KCl( aq ) l what mass of Pt(NH 3 ) 2 Cl 2 can be produced from 65 g of K 2 PtCl 4 ? l How much KCl will be produced? l How much from 65 grams of NH 3 ?

Limiting Reactant

1 2 5 ?=10 mol NH 3 3 2 9 ?=6 mol NH 3 15 (not available) H 2 is LR.

3 mol nitrogen 12 mol hydrogen 1 2 3 ?=6 mol NH 3 3 2 12 ?=8 mol NH 3 1 4 (not available) N 2 is LR.

Limiting Reactant l Reactant that determines the amount of product formed. l The one you run out of first. l Makes the least product. N 2 + H 2  NH 3  What mass of ammonia can be produced from a mixture of 100. g N 2 and 500. g H 2 ?  How much unreacted material remains?

1 2 3.57 ?=7.14 mol NH 3 3 2 248 ?=165.3 mol NH 3

Reaction 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. Percent Yield

68.5 kg CO is reacted with 8.60 kg H 2. Calculate the percent yield if the actual yield was 3.57x10 4 g CH 3 OH.

1 2445.6 ?=2445.6 mol CH 3 OH 2 1 4265.9 ?=2132.9 mol CH 3 OH

Examples l Aluminum burns in bromine producing aluminum bromide. In a laboratory 6.0 g of aluminum reacts with excess bromine. 50.3 g of aluminum bromide are produced. What are the three types of yield. 2 Al + 3 Br 2  2 AlBr 3

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?

Examples l 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 0.5065 g of brass is reacted with excess HCl, 0.0985 g of ZnCl 2 are eventually isolated. What is the composition of the brass (i.e. Zn% and Cu% )?

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