1. Stoichiometry Chapter 3 Chemical Formulas and Equations

2. Stoichiometry Chemical Equations Concise representations of chemical reactions

3. Stoichiometry Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

4. Stoichiometry Anatomy of a Chemical Equation Reactants appear on the left side of the equation. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

5. Stoichiometry Anatomy of a Chemical Equation Products appear on the right side of the equation. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

6. Stoichiometry Anatomy of a Chemical Equation The states of the reactants and products are written in parentheses to the right of each compound. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

7. Stoichiometry Anatomy of a Chemical Equation Coefficients are inserted to balance the equation. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

8. Stoichiometry Subscripts and Coefficients Give Different Information Subscripts tell the number of atoms of each element in a molecule

9. Stoichiometry Subscripts and Coefficients Give Different Information Subscripts tell the number of atoms of each element in a molecule Coefficients tell the number of molecules

10. Stoichiometry Cations = + (Lose Electrons) Anions = - (Gain Electrons)

11. Stoichiometry Trends in the Periodic Table Group 1 metals lose 1 electron to give 1+ cations, such as Na +1 or Li +1. Group 2 metals lose 2 electrons to give 2+ cations, such as Mg +2. Mg = 12 electrons 1s 2 2s 2 2p 6 3s 2

12. Stoichiometry Binary Ionic Compounds Compounds composed of two elements The total charges of both elements must be equal. First find each elements charge according to the periodic table. Mg = +2 (loses 2 electrons) N = -3 (gains 3 electron)

13. Stoichiometry Because compounds are electrically neutral: –The charge on the cation becomes the subscript on the anion. –The charge on the anion becomes the subscript on the cation. –Now we have 3 Mg with each +2 charge = +6 –And we have 2 N with each -3 charge = -6

14. Stoichiometry Na 1+ Cl 1- Ca 2+ Br 1- Al 3+ O 2-

15. Stoichiometry Using Chemical Formulas

16. Stoichiometry Using Formulas in Problem Solving Correctly written chemical formulas hold a large amount of information for the prepared student to find. The subscripts tell us the number of atoms of each kind that is present in the compound. Example: NaCl has one atom of sodium and one atom of chlorine. H 2 SO 4 has two atoms of hydrogen, one atom of sulfur and four atoms of oxygen

17. Stoichiometry The mass of a formula can be found by adding the masses of the atoms in the formula. The formula mass of any molecule is the sum of the average atomic masses of all atoms present in the formula. H 2 O H = 2 atoms (each with 1.01 amu) = 2.02 amu O = 1 atom (16 amu) = 16 amu Total = 16 +2.02 = 18.02 amu Formula Mass

18. Stoichiometry What is the formula mass for H 2 SO 4 ?

19. Stoichiometry Molar Mass = Formula Mass The sum of the atomic masses times the number of atoms of each kind of element is equal to the mass of one mole of the substance. Examples: Na = 23g/mol and Cl = 35.5g/mol so NaCl has a molar mass of 58.5g/mol Nitric acid is HNO 3 so its molar mass is H = 1 x 1 = 1 N = 14 x 1 = 14 O = 16 x 3 = 48 Total = 63g/mol

20. Stoichiometry What is the molar mass of Al 2 S 3 ?

21. Stoichiometry Molar Mass as a Conversion Factor To convert a known amount of a compound in moles to mass in grams, multiply the amount of moles by the molar mass. Amount of moles x molar mass (g/mol) = mass in grams Notice the moles cancel out

22. Stoichiometry What is the mass in grams of 2.5 moles of oxygen gas?

23. Stoichiometry To convert a known mass of a compound in grams to an amount in moles, the mass is divided by the molar mass. Mass in grams x 1 / molar mass (g/mol) = moles Notice the grams cancel out.

24. Stoichiometry Ibuprofen is C 13 H 18 O 2. What is the molar mass? If you have 33 g of ibuprofen, how many moles do you have?

25. Stoichiometry Percent Composition The percentage composition of each element in a compound can be determined using the formula and the atomic masses. Mass of element / mass of compound x 100 = % of element

26. Stoichiometry Example: Sodium chloride or NaCl Find the total mass of the compound Na = 23.0 Cl = 35.5 58.5g/mol Divide each mass by the total then x 100 % Na = 23.0 x 100 = 39.3%Na 58.5 % Cl = 35.5 x 100 = 60.7% Cl 58.5 Notice that the total of the percentages is always equal or very close to 100%.

27. Stoichiometry H2OH2O

28. Determining Chemical Formulas

29. Stoichiometry Calculating Empirical Formulas An empirical formula consists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole-number mole ratio of the different atoms in the compound. Begin by assuming you have 100g of substance and using the percent composition as the amount of grams. 78.1% of B = 78.1 grams of B 21.9% of H = 21.9 grams of H

30. Stoichiometry Once you have grams, convert to moles by looking up the amount of grams on the periodic chart which equals 1 mole. 78.1 grams B x 1 mole/10.81 g = 7.22 mol B 21.9 grams H x 1 mole/1.01 g = 21.7 mol H These values give us a mole ratio of 7.22 mol of B to 21.7 mol of H. However, we need this reduced to the smallest numbers. To find such ratio, divide each number of moles by the smallest number. 7.22/21.7 = 3.01 The ratio then is 1 mol B to 3 mol H Final Answer = BH 3

31. Stoichiometry A compound has 40% Fe and 60% S. Find the empirical formula

32. Stoichiometry Sometimes mass composition is known instead of percent composition. To find the empirical formula, convert mass composition to moles. Then calculate the smallest whole number mole ratio by dividing the smaller number over the larger. This removes 1 step!

33. Stoichiometry A 10.15 g sample of P and O has 4.433 g of P. What is the empirical formula? O = 4.433 g P = 10.15 – 4.433 = 5.717 g 4.433 g x 1 mole/16 g =.3573 mol Oxygen 5.717 g x 1 mole/30.97g =.1431 mol P.1431 /.3573 = 2.497 1 mol P = 2.5 mol O Must have whole numbers so multiply by 2 Final answer = P 2 O 5

34. Stoichiometry A 20 g compound has 4 g of Ca and some Br. What is the empirical formula?

35. Stoichiometry Calculating the molecular formula from the empirical formula The empirical formula is the lowest whole number ratio for a compound. It is not always the actual formula or the molecular formula. For example the empirical formula is CH but the molecular formula is C 2 H 4.

36. Stoichiometry To calculate the molecular formula from the empirical formula: x(empirical formula)=molecular formula x = experimental formula mass/empirical formula mass For example, the empirical formula is P 2 O 5 and it’s experimental molar mass is 283.89 g/mol. To find the empirical formula mass go to the periodic chart. P = 30.874 and O = 16 (30.874 x 2) + (16 x 5) = 141.94 amu 141.94 / 283.89 = 2.0001 2(P 2 O 5 ) = P 4 O 10

37. Stoichiometry What is the molecular formula of the compound with an empirical formula of CH and a formula mass of 78.11 amu?