ࡱ> ac` 9bjbj .V{{9     % &""",&.&.&.&.&.&.&$'G*R&" """R&g& ! ! !",& !",& ! !:`%,%@]P6 N% &}&0&%R* *%*%0"" !"""""R&R& !"""&""""*""""""""" : Molecular Formula Worksheet Molecular formula a formula showing the types and numbers of atoms combined in a single molecule of a molecular compound. It is a whole number multiple of the empirical formula. The relationship between a compounds empirical and molecular formula can be written as:  x(empirical formula) = molecular formula also x(empirical formula mass) = molecular formula mass 1. To determine the molecular formula of a compound, you must know the compounds molar mass. 2. Divide the molecular mass by the empirical formula mass to determine the whole number multiple (x). You may have to find the empirical formula in order to obtain the empirical formula mass. Problems: 1. In a previous problem, the empirical formula of a compound of phosphorus and oxygen was found to be P2O5. Experimentation shows that the molar mass of this compound is 283.89g/mol. What is the compounds molecular formula? 2. Determine the molecular formula of the compound with an empirical formula of CH and a molar mass of 78.110 g/mol. 3. A sample with a molar mass of 34.00 g/mol is found to consist of 0.44g H and 6.92g O. Find its molecular formula. 4. If 4.04g of N combine with 11.46g O to produce a compound with a molar mass of 108.0 g/mol, what is the molecular formula of this compound? 5. The empirical formula for trichloroisocyanuric acid, the active ingredient in many household bleaches, is OCNCl. The molar mass of this compound is 232.41g/mol. What is the molecular formula of trichloroisocyanuric acid. 6. The molar mass of a compound is 92g/mol. Analysis of a sample of the compound indicates that it contains 0.606g N and 1.390g O. Find its molecular formula. 7. Determine the molecular formula of a compound with an empirical formula of NH2 and a formula mass of 32.06 g/mol. 1. Empirical Formula = P2O5 Empirical Mass = 2(31.0) + 5(16.0) = 142g 283.89( 2142Molecular Mass = 283.89g.  2(P2O5) = P4O10  2. Empirical Formula = CH Empirical Mass = 1(12.0) + 1(1.0) = 13g 78.110( 613Molecular Mass = 78.110 amu  6(CH) = C6H6  3. 0.44g H1 mol H= 0.44 mol H0.44= 1 H1.0g H0.433 6.92g O1 mol O= 0.433 mol O0.433= 1 O16.0g O0.433 HO Empirical Formula = HO Empirical Mass = 1(1.0) + 1(16.0) = 17g 34.00( 217Molecular Mass = 34.00 amu  2(HO) = H2O2  4. 4.04 N1 mol N= 0.289 mol N0.289= 1 N14.0g N0.289 Xs 2 11.46g O1 mol O= 0.7163 mol O0.7163= 2.5 O16.0g O0.289 N2O5 Empirical Formula = N2O5 Empirical Mass = 2(14.0) + 5(16.0) = 108.0g 108.0= 1108.0Molecular Mass = 108.0 amu  1(N2O5) = N2O5  5. Empirical Formula = OCNCl Empirical Mass = 16.0+12.0+14.0+35.5 = 77.5g 232.41( 377.5Molecular Mass = 232.41 amu  3(OCNCl) = O3C3N3Cl3  6. 0.606g N1 mol N= 0.0433 mol N0.0433= 1 N14.0g N0.0433 1.390g O1 mol O= 0.08688 mol O0.08688= 2 O16.0g O0.0433 NO2 Empirical Formula = NO2 Empirical Mass = 1(14.0) + 2(16.0) = 46.0g 92= 246Molecular Mass = 92 g  2(NO2) = N2O4  7. Empirical Formula = NH2 Empirical Mass = 1(14.0) + 2(1.0) = 16g 32.06( 216Molecular Mass = 32.06 amu  2(NH2) = N2H4 ., - . / 0 1 y }   $ ( - A F U Z   < C (/VZ*+HMO]wxy|ʺʭhs-5B* CJ\ph jhs-CJ hs-CJH* hs-CJh(( hs-H* hs-5\hPjhs-CJUmHnHuhs-5>*\hs-@, . 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