1. Chapter Three Mass Relationships in Chemical Reactions

2. Chapter Three / Mass relationships in Chemical Reactions It possible to determine the empirical formula from the percentage of elements in the compound. 1- change % to g 2- change g to mole (remember the triangle). 3- divided by the smallest number of moles. 4- if there was fraction after division change to integer subscripts ( multiply by 1 or 2 or 3 etc until you reach integer. Percent Composition of a Compounds

3. Chapter Three / Mass relationships in Chemical Reactions Example 1: Determine the empirical formula of Vitamin C. it is compose of 40.92% of C, 4.58% of H, and 54.50% of O by mass? 1- we change from % to g 40.92 g of C, 4.58 g of H, 54.50 g of O 2- change from g to mole using Divided by the smallest number of mole which is 3.406 Percent Composition of a Compounds Mass (g) Molar massMole (mol) 40.92 12.011 = 3.407 mol of C n c = 4.58 1.008 = 4.54 mol of H n H = 54.50 16 = 3.406 mol of O n O =

4. Chapter Three / Mass relationships in Chemical Reactions 4- Because number of hydrogen is 1.33 then we start to multiply until we reach integer this is trail and error procedure: 1 x 1.33 = 1.33 2 x 1.33 =2.66 3 x 1.33 = 3.99 ≈ 4 Then we multiply all element with the same number which is 3 in this example. C : 1x3 =3, O: 1 x 3 = 3, H =4 Thus the empirical formula is C 3 H 4 O 3 Percent Composition of a Compounds 3.407 3.406 ≈ 1 C: 4.54 3.406 = 1.33 H: 3.406 = 1 O:

5. Chapter Three / Mass relationships in Chemical Reactions Percent Composition of a Compounds Example 2 : Allicin is the compound responsible of characteristic smell of garlic. an analysis of the compound gives the following percent composition by mass: C: 44.4 %, H: 6.21%, S: 39.5%, O: 9.86%. Calculate its empirical formula? 1- we change from % to g 44.4 g of C, 6.21 g of H, 39.5 g of S, 9.86 g of O. 2- change from g to mole using Divided by the smallest number of mole which is 0.62 Mass (g) Molar massMole (mol) 39.5 32.07 = 1.23 mol of S n S = 9.86 16 = 0.62 mol of O n O = 44.4 12.01 = 3.70 mol of C n C = 6.21 1.008 = 6.16 mol of H n H =

6. Chapter Three / Mass relationships in Chemical Reactions 3.70 0.62 ≈ 6 C: 6.16 0.62 ≈ 10 H: 0.62 = 1 O: 1.32 0.62 ≈ 2 S: Because all the numbers are integer then we do not need to do anything else and we just write the formula as following : The empirical formula is C 6 H 10 O S 2 Percent Composition of a Compounds