1. MATHEMATICS In Chemistry?!

2. You thought you were done with sig figs??? 1.Leading zeros are never significant 2.Imbedded zeros are always significant 3.Trailing zeros are only significant if a decimal point is specified 4.When multiplying/dividing- least total number of sig figs 5.When adding/subtracting- least number after the decimal

3. Practice: 100 100. 10.0 0.010 1010

4. The Mole 1 mole (mol) = 6.02 x 10 23 representative particles of that substance The SI unit for measuring the amount of a substance 6.02 x 10 23 = Avogadro’s number

5. Representative Particles The “species” that is present in a substance- usually atoms, molecules or formula units Fe, He, H 2, I 2, H 2 O, CaCl 2

6. Particles to Moles Conversion Factor: –Moles= particles x 1 mole 6.02 x 10 23 particles How many moles of magnesium is 1.25 x 10 23 atoms of magnesium?

7. Practice: Convert the following into moles 1)3.01 x 10 23 atoms of He 2)3.01 x 10 24 molecules of H 2 O 3)6.02 x 10 22 molecules of CO 2 4)9.03 x 10 23 molecules of NaOH 5)1.505 x 10 24 molecules of H 2 SO 4 6)1000 molecules of F 2

8. Moles to Particles Representative Particles = Moles x 6.02 x 10 23 particles 1 mole How many molecules are in 2.12 mol of propane (C 3 H 8 )? –How many atoms would that be?

9. Practice: Convert the following into atoms or molecules 1) 3.5 moles of CO 2 4) 8.2 moles of Ca(OH) 2 2) 1.25 moles of H 2 5) 7 moles of P 3)0.5 mole of K 6) 0.00025 mole of H 3 PO 4

10. The Mass of a Mole of an element Atomic masses- based on C-12 atom –carbon is 12x heavier than hydrogen The mass ratio of C atoms to H atoms remains the same no matter what unit is used to express the masses So 12.0 g of C atoms and 1.0 g of H atoms must contain the same number of atoms

11. Molar Mass The atomic mass of an element expressed in grams is the mass of a mole of the element. Molar Mass = the mass of 1 mol of atoms of any element C: 12.0 g O: 16.0 g O 2 : 32.0 g

12. gam= gram atomic mass

13. ALL MEAN THE SAME THING!!!!!! Gram atomic mass Molar mass Molecular weight Formula weight Gram formula weight

14. The Mass of a Mole of a Compound You can calculate the mass of a compound by adding the atomic masses of the atoms that make it up. S- 32.1 g (1)= 32.1 g/mol O- 16.0 g (3)= + 48.0 g/mol 80.1 g/mol

15. Calculate the Molar Mass of a Compound: Find the molar mass of: H 2 O and H 2 O 2

16. Be careful! What is the molar mass of oxygen? –Depends on what you think the representative particle is- O or O 2

17. How would you measure 3.00 mol of NaCl? Use the molar mass of an element or compound to convert between the mass of a substance and the # of moles. Mass (g) = # of moles x mass (g) 1 mol

18. NaCl Molar Mass of NaCl? –58.5 g/mol 3.00 mol x 58.5 g 1 mol =176 grams

19. You can also do the reverse: Moles= mass (g) x 1 mole mass (g) Na 2 SO 4 - molar mass = 142.1 g/mol 10.0 g x 1 mol 142.1 g = 7.04 x 10 -2 mol

20. Mole-Volume Relationships AVOGADRO’S HYPOTHESIS: equal volumes of GASES at the same temperature and pressure contain the same number of particles.

21. How can that be? The particles are not the same size, but the particles are so far apart (gases) that a collection of relatively large particles does not require much more space than the same number of relatively small particles.

22. The Volume of a gas Varies with temperature and pressure. Because of these variations, the volume of a gas is usually measured at Standard Temperature and Pressure (STP)

23. STP 0°C (273 K) 101.3 kPa = 1 atmosphere (atm)

24. At STP: 1 MOL (6.02 X 10 23 particles) OF ANY GAS OCCUPIES A VOLUME OF 22.4 LITERS. 22.4 L = molar volume of a gas (at STP)

25. The Molar Volume Is used to convert a known number of moles of gas to volume of the gas at STP Volume of gas= moles of gas x 22.4 L 1 mol

26. EXAMPLE 0.375 mol O 2 (at STP) 0.375 mol x 22.4 L 1 mol = 8.40 L

27. Now for the reverse Moles of gas= volume of gas (L) x 1 mol 22.4 L

28. Percent Composition and Chemical Formulas The relative amounts of components of a mixture or compound are expressed as the % composition or the % by mass of each element in the compound.

29. Percent by mass The percent by mass of an element in a compound is the number of grams of the element divided by the mass in grams of the compound, multiplied by 100%. % mass of element= mass of element x 100% mass of compound

30. Percent Composition from Mass Data When a 13.60 g sample of a compound containing only Mg and O is decomposed, 5.40 g of oxygen is obtained. What is the percent composition of this compound?

31. The Math % Mg= 8.20 g x 100% 13.60 g –= 60.3% Mg % O= 5.40 g x 100% 13.60 g –= 39.7% O

32. Percent Composition from the Chemical Formula The subscripts in the formula of the compound are used to calculate the mass of each element in a mole of that compound. The sum of these masses is the Molar Mass.

33. So, Using the individual masses of the elements and the molar mass, you can calculate the % by mass of each element in one mole of the compound. % mass= Mass of element in 1 mol of compound x 100% molar mass of compound

34. EXAMPLE Calculate the % composition of propane (C 3 H 8 ) Mass of C in 1 mol= 36.0 g Mass of H in 1 mol= 8.0 g Molar mass= 44.0 g/mol % C= 36.0 g x 100% 44.0 g = 81.8% carbon % H= 8.00 g x 100% 44.0 g = 18% hydrogen

35. Practice Problem- % Composition Calculate the percent composition for copper (II) nitrate from its formula.

36. Practice Problem- % Composition From laboratory measurements, a sample of a pure compound is known to have a mass of 3.74 grams. Analysis of the sample shows 1.10 g of calcium, 0.880 g sulfur, and 1.76 g oxygen. What is the percent composition of this compound?

37. % Composition and HYDRATES As some compounds crystallize from a water solution, they trap water molecules- HYDRATES. Find the percent of water in Na 2 CO 3. 10 H 2 O

38. Percent Composition as a Conversion Factor You can use % composition to calculate the number of grams of any element in a specific mass of a compound. –Multiply the mass of the compound by a conversion factor based on the % composition of the element in the compound.

39. EXAMPLE You can use the ratio 81.8 g C/ 100 g C 3 H 8 to calculate the mass of carbon contained in 82.0 g C 3 H 8. –82.0 g C 3 H 8 x 81.8 g C 100g C 3 H 8 = 67.1 g C

40. Empirical Formula Basic ratio; gives the lowest whole- number ratio of the atoms of the elements in a compound. An empirical formula may or may not be the same as a molecular formula. –HO H 2 O 2

41. Practice Problem- Empirical Formulas The percent composition by mass of a compound is 56.6% potassium, 8.7% carbon, 34.7% oxygen. Find its empirical formula. Assume that there are 100 grams of the compound. Then calculate the number of moles of atoms of each element in the sample.

42. Practice Problem- Empirical Formula Analysis shows that a compound contains 26.56% K, 35.41% Cr, and 38.03% O. Find the simplest formula for this compound.

43. Practice Problem- Empirical Formula Analysis of a 10.150 g sample of a compound known to contain only phosphorous and oxygen yields 5.717 g of oxygen. What is the empirical formula of this compound?

44. Molecular Formula The molecular formula tells the actual number of each kind of atom present in a molecule of a compound.

45. The Molecular Formula Of a compound is either the same as its experimentally determined empirical formula, or it is a simple whole-number multiple of its empirical formula.

46. Example Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH 4 N

47. The steps: 1.Calculate the empirical formula mass. 2.Divide the molar mass by the empirical formula mass to obtain a whole number. 3.Multiply the formula subscripts by this value to get the molecular formula. Empirical Formula efmMM/efmMolecular Formula CH 4 N30.02C2H8N2C2H8N2

48. Practice Problem- Molecular Formula The empirical formula of a compound was found to be P 2 O 5. Experiments show that the formula mass of this compound is 283.889 g. What is the molecular formula of this compound? What is the name?

49. More molecular formula problems 1.Determine the molecular formula of a compound having an empirical formula of CH and a formula mass of 78.110 g. 2.A compound with a formula mass of 42.0 g is found to be 85.64% C and 14.36% H by mass. Find its molecular formula.