Percent Composition and Empirical Formulas in C#

Read ECC200 in C# Percent Composition and Empirical Formulas

If the formula of a compound is known, it is a fairly straightforward task to determine the percent composition of each element in the compound For example, suppose you want to calculate the percentage hydrogen and oxygen in water, H2O First calculate the molecular mass of water: 1 mol H2O = 2 mol H + 1 mol O Substituting the masses involved: 1 mol H2O = 2 (10079 g/mol) + 1600 g/mol = 180158 g/mol (intermediate calculation don t worry about significant figures yet) percentage hydrogen = [mass H/mass H2O] 100 = [2(10079 g/mol)/180158 g/mol] 100 = 1119% H = [mass O/mass H2O] 100 = [1600 g/mol/180158 g/mol] 100 = 8881% O

As a good check, add the percentages together They should equal to 100% or be very close Determine the mass percent of each of the elements in C6H12O6 (FM) = 180158 amu Answer: (6C atoms)(12011 amu/atom) 100% = 40002% (180158 amu) m (12 Hatoms)(1008 amu/atom ) 100% = 6714% %H = (180158 amu) (6Oatoms)(159994 amu/atom) 100% = 532846% %O = (180158 amu) Total = 100001% %C = The total is a check It should be very close to 100% In the problems above, the percentage data was calculated from the chemical formula, but the empirical formula can be determined if the percent compositions of the various elements are known The empirical formula tells us what elements are present in the compound and the simplest whole-number ratio of elements The data may be in terms of percentage, or mass, or even moles But the procedure is still the same: convert each to moles, divide each by the smallest number, then use an appropriate multiplier if needed The empirical formula mass can then be calculated If the actual molecular mass is known, dividing the molecular mass by the empirical formula mass gives an integer (rounded if needed) that is used to multiply each of the subscripts in the empirical formula This gives the molecular (actual) formula, which tells which elements are in the compound and the actual number of each For example, a sample of a gas was analyzed and found to contain 234 g of nitrogen and 534 g of oxygen The molar mass of the gas was determined to be about 90 g/mol What are the empirical and molecular formulas of this gas Answer: 1molN ( 2 34 g N) = 0 167 mol N 140 g N 1molO (534 g O) = 0 334 mol O 160 g O Empirical Formula = NO2 0167 =1 N 0167 0334 = 2 O 0167 Formula mass

The molecular formula may be determined by dividing the actual molar mass of the compound by the empirical molar mass In this case the empirical molar mass is 46 g/mol 90 g/mol = 1 96 which, to one significant figure, is 2 Therefore, the molecular Thus 46 g/mol formula is twice the empirical formula N2O4

Be sure to use as many significant digits as possible in the molar masses Failure to do so may give you erroneous ratio and empirical formulas

As we have discussed previously, the balanced chemical equation not only indicates which chemical species are the reactants and the products, but also indicates the relative ratio of reactants and products Consider the balanced equation of the Haber process for the production of ammonia: N2(g) + 3H2(g) 2 NH3(g) This balanced equation can be read as: 1 nitrogen molecule reacts with 3 hydrogen molecules to produce 2 ammonia molecules But as indicated previously, the coefficients can stand not only for the number of atoms or molecules (microscopic level), they can also stand for the number of moles of reactants or products The equation can also be read as: 1 mol of nitrogen molecules reacts with 3 mol of hydrogen molecules to produce 2 mol of ammonia molecules And if the number of moles is known, the number of grams or molecules can be calculated This is stoichiometry, the calculation of the amount (mass, moles, particles) of one substance in a chemical reaction through the use of another The coefficients in a balanced chemical equation define the mathematical relationship between the reactants and products, and allow the conversion from moles of one chemical species in the reaction to another Consider the Haber process above How many moles of ammonia could be produced from the reaction of 200 mol of nitrogen with excess hydrogen