Download presentation

Presentation is loading. Please wait.

Published byEvelyn MacLeod Modified over 3 years ago

1
The MOLE Introduction to stoichiometry

2
AVOGADRO S NUMBER How would you find the mass of an object with too little mass to register on your balance?

3
AVOGADRO'S Number If you could put a large number of the same object on the balance, you could get a measurable mass. Then, by dividing this measurable mass by the number of objects, you would get the mass of one individual object. This idea occurred to Avogadro, an Italian scientist who did not know how large the number would be, but who hypothesized

4
That once one combining mass of an element was known, it must have the Same number of atoms as one combining mass of a different element. By the end of the nineteenth century, the SIZE of that number had been determined And a word, meaning a heap had been invented as the unit for that number. AVOGADRO S NUMBER: 6.02 x = 1 mole

5
The natural substance with the LEAST MASS is H 2, hydrogen, with a mass of atomic mass units. Recall that each H atom has atomic mass units. By definition, 1 mole of H 2, hydrogen, has a mass of g (usually rounded to 2.02 g). How many grams is 1 molecule of hydrogen gas?

6
Finding the Mass in grams of one molecule of H 2 and 1 atom of H: = = = Since molecules are the representative particles of hydrogen gas

7
Review What number represents a mole? Avogadro s number Gram formula mass divided by the Avogadro number = representative particle mass in grams. A representative particle mass will have an exponent in the g. range (possibly as big as g)

8
REVIEW KINDS OF REPRESENTATIVE PARTICLES: ions, atoms, molecules, formula units. So if we are referring to particles, or larger number of particles, we are talking about or larger number of moles. The least number of grams of a naturally occurring substance is 2.02 grams/mole, which is H 2.

9
Problem solving in quantitative chemical problems. Must use common sense; troubleshoot obvious errors by making sure that inverses have not been used as conversion factors. EXPECT information to be missing from problems: Look for missing info in the text, other problems or sample problems, previous chapters, tables in same or previous chapters, and reference tables including of course the Periodic Table.

10
For example in practice prob. 1 Find data about apples per bushel. Found on p 172: 1 doz apples/.20 bushel 12 apples / dozen 12 apples/.20 bushels = x apples/.50 bushels. The question asks about the 2 mass of.50 bushel. Mass of 1 apple, found in sample prob 7.1. Number of apples -> mass of apples. 2.0 kg / 12 apples 12ap/.20 bu=x ap/.50 bu X ap=12ap(.5bu) /.2bu X =30 ap 30 ap/y kb = 12 ap/2.0 kg 12 ap(y kg)= 30 ap(2 kg)

11
Sample prob x atoms of Mg. How many moles is this? 6.02 x rep particles (atoms)/ 1 mol x /x moles = 6.02 x /1 mole X moles =1.25 x / 6.02 x The correct answer MUST be converted into _____/1, the 1 then is omitted!

12
2.12 moles of C 3 H 8 contains X atoms. 1 mole contains a certain number of atoms, which is 6.02 x f the number of atoms in one individual molecule. How many atoms per 1 molecule? (11 atoms/ 1molecule) (6.02x atoms/1mole molecules)(2.12 mole) are present) =# of atoms present = ___?_atoms YOU MUST show the units that cancel before you cancel them!

13
2.12 moles of C 3 H 8 contains x atoms. 1 mole contains a certain number of atoms, which is 6.02 x f the number of atoms in one individual molecule. How many atoms per 1 molecule?

14
Gather all given & known information: Atoms is the final quantity desired. = x atoms Cancel all possible units, checking to make sure that appropriate units cancel rather than multiply! Evaluate the answer including the remaining unit: means give the numerical answer in decimal form!

15
Frequently, lab measurable quantities are given or wanted. ONLY use 6.02 x as a factor if a representative particle is a known or wanted quantity. IF grams are wanted and moles are given, use the gram formula mass ( gfm ) as the conversion factor. For example, how many moles of pentene are found in 65.0 g. of pentene, C 5 H 10. What is pentene? It is the compound whose formula is next to the name. What do we do next?

16
Solving a gram to mole problem: 1) find the gram formula mass (gfm) of the substance in question 2) by definition, the mass of one mole is the gfm. 3) consult the PERIODIC TABLE for average atomic masses. Remember that the entire atomic mass is made of both protons & neutrons ; DON T choose the atomic number! 4) Find the total mass of each element; find the sum of all the masses.

17
For example, C 5 H 10 Contains 5 Carbons and 10 hydrogens. The atomic number of C is 6; its average atomic mass is The atomic number of H is 1; its average atomic mass is The total gram mass of C in 1 mole of compound is 5 x 12.0g= 60.0 g The total gram mass of H in the compound is 10 x 1.01 gram=10.1 g. The sum of all the elements in the compound is thus 70.1 g/mole. = =0.927 moles

18
How many moles is 65.0 g of (NH 4 ) 3 (PO 4 )? Steps: gram formula mass = mass/1 mole. Total number of N in this formula? 3 x 1 = 3 Total number of H in this formula? 3x4 = 12 Total # of P=1; of O=4 Gram atomic masses of N, H, P and O? N : 7 or 14.0? H : 1 or 1.01? P : 15 or 30 or 31.0? O : 8 or 15 or 16.0? Answers: N, 14.0g; H, 1.01g, P, 31.0g; O, 16.0 g.

19
The gram formula mass of ammonium phosphate is Then we can find the number of moles of ammonium phosphate in 65.0 g. By ….. 3 x 14.0g = 42.0 g.N 12 x 1.01g = 12.1 g H 1 x 31.0 g = 31.0 g P 4 x 16.0 g = 64.0 g O _________________ Gfm=

20
Fill in the blanks: Grams of ammonium phosphate available = ________ 1mole/Molar mass (gfm) of ammonium phosphate = ______/_____ ( _______) (__________) = ___ moles ( )

21
Answer: Grams available: 65.0 g Molar mass: Invert the molar mass so that g will cancel out! =0.436 moles (NH 4 ) 3 PO 4

22
And the answer is =0.436 g (NH 4 ) 3 PO 4

23
The reverse procedure gives grams when moles are given. How many grams are there in moles of O 2 ? We know the number of moles. We go to the periodic table to find the gfm of oxygen gas. Since moles must be cancelled, the g/1mol format is NOT inverted: The only question left is, what is the CORRECT formula mass of oxygen gas. 1 atom of O is amu; 1 mol of O is g and one mole of O 2 = 32.0 g.

24
To set up the problem, moles O 2 = ____ g O 2 And the answer is ………… g O 2

25
For gases at standard temperature of 0 o C and 1 atm. 1 mole of gas occupies 22.4 L. This conversion factor does not apply to liquids or solids which have definite density. The density of a gas varies with pressure and temperature. The abbreviation for 0 o C and 1 atm. is S.T.P.

26
Here are some molar volume questions: If you have 11.2 L of H 2 at STP, how many moles and how many grams of hydrogen gas are present. Since one mole of H 2 = 2.02 g, And there are moles There are 1.01 g. Set this entire mathematical scenario up using the format previously used! 11.2 L = mol

27
More molar volume questions 1. How many grams of oxygen gas are present in 3.54 L at STP? 2. How many grams of iron are present in 3.54 L at STP? 3. How many grams of C 2 H 4 (ethene gas) at STP are present in 2.0 L? 4. How many grams of CH 4 (natural gas or methane) are present in 4.48 L at STP?

28
Did you notice the trick question? Which one is it? At 0 o C, one of those substances is NOT A GAS. The solid substance does not follow the molar volume law. You would have to know its density in order to solve problem 2. The other problems simply rely on

29
and Gfm of O 2 x # of moles And similar action with problems 3 & 4. You are REQUIRED to have solutions to ALL these problems IN YOUR NOTES, including #2, (which requires looking up and using the density of Fe!)

30
% composition and moles Refer to earlier notes about the % composition of water Dalton & atomshttp://www.newmex.com/lpchemistry 88.9% of water is oxygen by mass, while 11.1% of water is hydrogen by mass. This means that out of every 100 g, ___ g is oxygen. How many moles of oxygen would there be in 100 g of water?

31
Using percent composition. What is the ratio of moles of H to moles of O? Note: O, not O 2 !

32
Finding empirical formulas H 11.1 O 5.5 reduces to H 2 O 1 or our familiar H2OH2O The same process can be used with any compound for which the % composition has been determined by experiment. For example, a compound exists which is 75% C and 25% H by mass. It is NOT an acid, meaning that __ comes ____.

33
What is the formula for this compound? Check out the flowchart for naming of compounds. Your text gives P x Q y as a template for a binary compound. If the compound is an acid, P is H, hydrogen. Since this compound is NOT an acid, P must be __ C. Change 75% into 75 g C and determine the number of moles present.

34
Fill in the blanks in each step for finding the formula. The first element in the formula is C. The second element in the formula is H, by default.

35
Insert the number of moles you have found in the subscript variables. C --- H ---- Divide the smaller subscript into the larger subscript. In this case, the rounded integers will be even. C 6.25 H /6.25 ~ 4 C 1 H 4 or CH 4 The substance is methane.

36
A more interesting compound is composed of 69.9% of a compound is iron. The remainder is oxygen. What % of this compound is oxygen? Change 69.9% into ___g/100g iron, and find the # of moles of iron. Since iron is a ____, it appears __ in the formula. Find the # of moles of oxygen.

37
1.25 moles of Fe are present for every 1.88 moles of O. Dividing 1.88/1.25 yields a ratio of 1.5/1 of O. Fe 1 O 1.5 is the result. But subscripts must be integers. Multiplying each subscript by 2 yields the correct empirical formula, Fe 2 O 3. 3:4 or 4:3 ratios are also possible.

38
If there are 3 or more elements, a polyatomic ion may be involved. N 2 O 6 must be reduced to (NO 3 ) 2 and N 3 H 12 would reduce to ? (NH 4 ) 3. Obviously, the more adequately you have memorized your polyatomic ions, the easier this part will be for you!

39
The situation with molecules Molecular compounds are often made up of apparently repeating units. For example, there are many many molecular compounds with the empirical formula C n H 2n. Examples would include C 2 H 4 and C 30 H 60. Clearly, the molar mass of the first compound would be 28g/mol, while the molar mass of the second would be 420 g/mol. Note that instead of being reduced, these molecular formulas are

40
Expanded. If you found that the EMPIRICAL FORMULA for a molecule was C n H 2n, you would have to know its molecular mass in order to provide its molecular formula. If the molar mass were 420 g/mol, and you divided the unit mass of 14 g/mol (for C 1 H 2 ) into 420, you would find that there were in fact 30C : 60 H. Thus, a molecular formula can be determined from an empirical formula IFF the molar mass can be determined.

41
Have we already learned a way to determine molar mass … at least for one set of compounds???? The answer is Yes, for gases at S.T.P. Remember, at S.T.P. (__atm pressure and __ o C), _____ L of ANY gas contain 1 mole of molecules. This number of liters can be massed. A fractional number of liters represents the same fraction of the number of moles and

42
thus the same fraction of the molar mass. If the % composition of a compound HAS BEEN PREVIOUSLY DETERMINED AND This substance can be obtained in the gaseous state at standard conditions, Then its molecular formula can be determined. This was the work of chemists following the acceptance of Dalton s theory.

43
The end: return to Mrs. Ditkowsky s Home Page

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google