Download presentation

1
**% Composition, Empirical and Molecular Formulas**

Chemical Quantities The Mole, % Composition, Empirical and Molecular Formulas

2
**How you measure how much?**

You can measure mass, or volume, or you can count pieces. We measure mass in grams. We measure volume in liters. We count pieces in MOLES.

3
**Moles Defined as the number of carbon atoms in exactly**

12 grams of carbon-12. 1 mole is 6.02 x particles. Treat it like a very large dozen 6.02 x is called Avagadro’s number.

4
**Representative particles**

The smallest pieces of a substance. For a molecular compound it is a molecule. For an ionic compound it is a formula unit. For an element it is an atom.

5
**Types of questions 3 12 3 2 5 How many oxygen atoms in the following?**

CaCO3 Al2(SO4)3 How many ions in the following? CaCl2 NaOH 3 12 3 2 5

6
**Types of questions using the equality; 1 mole = 6.02 x 1023**

How many molecules of CO2 are the in 4.56 moles of CO2 ? 4.56 mole x x1023 mc = mole How many moles of water is 5.87 x molecules? 5.87 x mc x mole = x1023 mc 2.75x1024mc mole

7
**Types of questions using the equality; 1 mole = 6.02 x 1023**

How many atoms of carbon are there in 1.23 moles of C6H12O6 ? 1.23 moles x x1023 mc x 6 atoms = mole mc How many moles is 7.78 x 1024 formula units of MgCl2? 7.78x1024 FU x 1 mole = mole x1024 FU 4.44x1024 atoms

8
**Measuring Moles Remember relative atomic mass?**

The amu was one twelfth the mass of a carbon 12 atom. Since the mole is the number of atoms in 12 grams of carbon-12, the decimal number on the periodic table is also the mass of 1 mole of those atoms in grams.

9
**Gram Atomic Mass The mass of 1 mole of an element in grams.**

12.01 grams of carbon has the same number of pieces as grams of hydrogen and grams of iron. We can right this as g C = 1 mole We can count things by weighing them.

10
**Examples How much would 2.34 moles of carbon weigh?**

2.34 moles C x 12 g C mole How many moles of magnesium in g of Mg? 24.31 g Mg x mole = g Mg = g 1.013 mole

11
**How many atoms of lithium in 1.00 g of Li?**

1.00 g Li x 1 mole x 6.02x1023 atoms g Li mole How much would 3.45 x 1022 atoms of U weigh? 3.45x1022 atoms U x 1 mole x g U x1023atoms 1 mole 8.60x1022 atoms 13.6 g

12
What about compounds? in 1 mole of H2O molecules there are two moles of H atoms and 1 mole of O atoms To find the mass of one mole of a compound determine the moles of the elements they have Find out how much they would weigh add them up

13
**What about compounds? What is the mass of one mole of CH4?**

1 mole of C = 12 g 4 mole of H x 1 g = 4g 1 mole CH4 = = 16g The Gram Molecular mass of CH4 is 16.05g The mass of one mole of a molecular compound.

14
**Gram Formula Mass The mass of one mole of an ionic compound.**

Calculated the same way. What is the GFM of Fe2O3? 2 moles of Fe x 56 g = 112 g 3 moles of O x 16 g = 48 g The GFM = 112 g + 48 g = 160g

15
**Molar Mass The generic term for the mass of one mole.**

The same as gram molecular mass, gram formula mass, and gram atomic mass.

16
Examples Calculate the molar mass of the following and indicate what type it is. Na2S 2 (23) + 32 = N2O4 2(14) + 4(16) = C = 78g 1mole Na2S = 78g Gram Formula Mass = 92g 1 mole N2O4 = 92g Gram Molecular Mass 12g 1 mole C = 12 g Gram Atomic Mass

17
**Molar Mass Cont. Ca(NO3)2 40 + 2(14) + 6(16) = 40 + 28 + 96 = 164g**

1 mole Ca(NO3)2 = 164g C6H12O6 6(12) + 12(1) + 6(16) = = 180g 1 mole C6H12)6 = 180g Gram Molecular Mass (NH4)3PO4 3(14) + 12(1) (16) = = 149g 1 mole (NH4)3PO4 = 149g Gram Formula Mass Gram Formula Mass

18
**Finding moles of compounds Counting pieces by weighing**

Using Molar Mass Finding moles of compounds Counting pieces by weighing

19
**Molar Mass The number of grams of 1 mole of atoms, ions, or molecules.**

We can make conversion factors from these. To change grams of a compound to moles of a compound.

20
For example How many moles is 5.69 g of NaOH?

21
For example How many moles is 5.69 g of NaOH?

22
**For example How many moles is 5.69 g of NaOH?**

need to change grams to moles

23
**For example How many moles is 5.69 g of NaOH?**

need to change grams to moles for NaOH

24
**For example How many moles is 5.69 g of NaOH?**

need to change grams to moles for NaOH 1mole Na = 23g 1 mol O = g 1 mole of H = 1 g

25
**For example How many moles is 5.69 g of NaOH?**

need to change grams to moles for NaOH 1mole Na = 23g 1 mol O = 16 g mole of H = 1 g 1 mole NaOH = 40 g

26
**For example How many moles is 5.69 g of NaOH?**

need to change grams to moles for NaOH 1mole Na = 23g 1 mol O = 16 g 1 mole of H = 1 g 1 mole NaOH = 40 g

27
**For example How many moles is 5.69 g of NaOH?**

need to change grams to moles for NaOH 1mole Na = 23g 1 mol O = 16 g mole of H = 1 g 1 mole NaOH = 40 g

28
**Examples How many moles is 4.56 g of CO2? 4.56g CO2 x 1 mole 1 44gCO2**

How many grams is 9.87 moles of H2O? 9.87 moles x 18g H2O mole = moles = 178g

29
**Examples How many molecules in 6.8 g of CH4?**

6.8g CH4 x 1 mole x 6.02x1023 mc g CH mole 49 molecules of C6H12O6 weighs how much? 49 mc x 1 mole x g C6H12O6 = x1023 mc mole 8820 g____= 1.5x10-20 g 6.02x1023 = 2.56x1023 mc

30
Gases and the Mole

31
**Gases Many of the chemicals we deal with are gases.**

They are difficult to weigh. Need to know how many moles of gas we have. Two things effect the volume of a gas Temperature and pressure Scientists compare gases at Standard Temperature and Pressure

32
**Standard Temperature and Pressure**

0ºC and 1 atm pressure abbreviated STP At STP 1 mole of gas occupies 22.4 L Called the molar volume Avogadro’s Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles.

33
**Examples What is the volume of 4.59 mole of CO2 gas at STP?**

4.59mole x 22.4L mole How many moles is L of O2 at STP? 5.67L x 1 mole L What is the volume of 8.80g of CH4 gas at STP? 8.80g CH4 x 1mole x 22.4L g CH mole = = 103L = .523moles = = 12.3L

34
**We have learned how to change moles to grams moles to atoms**

moles to formula units moles to molecules moles to liters molecules to atoms formula units to atoms formula units to ions

35
Mass Periodic Table Moles

36
Mass Volume Periodic Table Moles

37
Mass Volume Periodic Table 22.4 L Moles

38
Mass Volume Periodic Table 22.4 L Moles Representative Particles

39
**Mass Volume Moles 6.02 x 1023 Representative Particles 22.4 L**

Periodic Table 22.4 L Moles 6.02 x 1023 Representative Particles

40
**Mass Volume Moles 6.02 x 1023 Representative Particles Atoms 22.4 L**

Periodic Table Moles 6.02 x 1023 Representative Particles Atoms

41
**Mass Volume Moles 6.02 x 1023 Representative Particles Ions Atoms**

Periodic Table Moles 6.02 x 1023 Representative Particles Ions Atoms

42
**Percent Composition Like all percents Part x 100 % whole**

Find the mass of each component, divide by the total mass.

43
Example Calculate the percent composition of each element in a compound that is 29.0 g of Ag with 4.30 g of S. Ag g S g 33.3g /33.3 = x 100 = 87.09% /33.3 = x 100 = 12.91%

44
**Getting % from the formula**

If we know the formula, assume you have 1 mole. Then you know the pieces and the whole.

45
**Examples Calculate the percent composition of C2H4? C 2(12g)=24**

H 4(1g) = +4 28g /28 = x 100 = 85.71% /28 = x 100 = 14.29%

46
**Example Calculate the percent composition of Aluminum carbonate.**

Al2(CO3)3 Al 2(27g)= 54 C 3(12g)= 36 O 9(16)= 144 234g /234 = x 100 = 23.08% /234 = x 100 = 15.38% /234 = x 100 = 61.54%

47
**• Step 2: Multiply the elements %, by the mass of the compound given.**

You can also calculate the mass of an element in a given amount of a compound using % composition. • Step 1: calculate the % comp. only of the element you want to find the mass of. • Step 2: Multiply the elements %, by the mass of the compound given. Example: Calculate the mass of sulfur in 3.54g of H2S. MM of H2S = H 2 (1) = S 1(32) = +32 34g H2S % S = 32/34 x 100 = 94.1% S 94.1% x 3.54g = 3.33g S

48
**Calculate the mass of nitrogen in 25g of (NH4)2CO3.**

H 8(1g) = C 1(12g) = O 8(16g) = +128 176g (NH4)2CO3 %N = 28/176 x 100 = 15.91% 15.91% x 25g = 4.0g N

49
**Calculate the mass of magnesium in 97.4g of Mg(OH)2.**

Mg 1 (24g) = 24 O 2(16g) = 32 H 2 ( 1g) = +2 58g Mg(OH)2 %Mg = 24/58 x 100 = 41.38% 41.38% x 97.4g = 40.3g Mg

50
**From percentage to formula**

Empirical Formula From percentage to formula

51
The Empirical Formula The lowest whole number ratio of elements in a compound. The molecular formula the actual ration of elements in a compound. The two can be the same. CH2 empirical formula C2H4 molecular formula C3H6 molecular formula H2O both

52
**Calculating Empirical**

Just find the lowest whole number ratio C6H12O6 CH4N It is not just the ratio of atoms, it is also the ratio of moles of atoms. In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen. In one molecule of CO2 there is 1 atom of C and 2 atoms of O.

53
**Calculating Empirical**

Means we can get ratio from percent composition. Assume you have a 100 g. The percentages become grams. Can turn grams to moles. Find lowest whole number ratio by dividing by the smallest moles.

54
Example Calculate the empirical formula of a compound composed of % C, % H, and %N. Assume 100 g so 38.67 g C x 1mol C = mole C g C 16.22 g H x 1mol H = mole H g H 45.11 g N x 1mol N = mole N g N

55
**Example The ratio is 3.223 mol C = 1 mol C 3.222 mol N 1 mol N**

The ratio is mol H = 5 mol H mol N mol N C1H5N1

56
**A compound is 43.64 % P and 56.36 % O. What is the empirical formula?**

43.64 g P x 1mol P = mole P g P 56.36 g O x 1mol O = mole O g O The ratio is mol O = 2.5 mol O mol P mol P Can not have 2.5 atoms! Double P2O5

57
**Caffeine is 49. 48% C, 5. 15% H, 28. 87% N and 16. 49% O**

Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula? 49.48 g C x 1mol C = mole C g C 5.15 g H x 1mol H = 5.15 mole H g H 28.87 g N x 1mol N = mole N g N 16.49 g O x 1mol O = mole O g O

58
**The ratio is 4.123 mol C = 4 mol C 1.031 mol O 1 mol O**

The ratio is 5.15 mol H = 5 mol H mol O mol O The ratio is mol N = 2 mol N mol O mol O C4H5N2O1

59
**Empirical to molecular**

Since the empirical formula is the lowest ratio the actual molecule would weigh more. By a whole number multiple. Divide the actual molar mass by the the mass of one mole of the empirical formula. Caffeine has a molar mass of 194 g. what is its molecular formula? C4H5N2O1 = 97g 194/97 = 2 Molecular Formula = C8H10N4O2

60
**Example 71.65 g Cl x 1mol Cl = 2.047 mole Cl 35 g Cl**

A compound is known to be composed of % Cl, 24.27% C and 4.07% H. Its molar mas is known (from gas density) is known to be g. What is its molecular formula? 71.65 g Cl x 1mol Cl = mole Cl g Cl 24.27 g C x 1mol C = mole C g C 4.07 g H x 1mol H = 4.07 mole H g H

61
**The ratio is 2.047 mol Cl = 1 mol Cl 2.023 mol C 1 mol C**

The ratio is 4.07 mol H = 4 mol H mol C mol C Empirical Formula = CClH4 = 51g 98.96/51 = 2 Molecular Formula = C2Cl2H8

Similar presentations

© 2020 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google