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Chapter Three Mass Relationships in Chemical Reactions Chapter Three / Mass relationships in Chemical Reactions Percent Composition of a Compounds • 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. Chapter Three / Mass relationships in Chemical Reactions Percent Composition of a Compounds 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 Mass (g) 40.92 g of C, 4.58 g of H , 54.50 g of O Mole Molar mass 2- change from g to mole using (mol) 40.92 nc = 12.011 = 3.407 mol of C 4.58 nH = 1.008 = 4.54 mol of H 54.50 nO = = 3.406 mol of O 16 Divided by the smallest number of mole which is 3.406 Chapter Three / Mass relationships in Chemical Reactions Percent Composition of a Compounds C: 3.407 3.406 ≈1 H: 4.54 3.406 = 1.33 O: 3.406 3.406 =1 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 C3H4O3 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 Mass (g) 44.4 g of C, 6.21 g of H , 39.5 g of S, 9.86 g of O. Mole Molar mass (mol) 2- change from g to mole using nC = 44.4 12.01 = 3.70 mol of C nH = 6.21 1.008 = 6.16 mol of H 39.5 9.86 nO = = 1.23 mol of S = 0.62 mol of O 32.07 16 Divided by the smallest number of mole which is 0.62 nS = Chapter Three / Mass relationships in Chemical Reactions Percent Composition of a Compounds 3.70 6.16 H: ≈6 ≈ 10 0.62 0.62 1.32 0.62 O: S: ≈ 2 =1 0.62 0.62 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 C6H10O S2 C: