EXERCISE 10 (continued) NAME 6. An organic compound is analyzed and found to be

Question

EXERCISE 10 (continued) NAME 6. An organic compound is analyzed and found to be carbon 51.90%, hydrogen 9.80% and chlorine 38.30%. What is the empirical formula of this compound? sample of oxygen gas, Os, weighs 28.4 g. How many molecules of O2 and how many atoms of O are present in this sample? 7. A molecules of O2 atoms of O 8. A mixture of sand and salt is found to be 48 percent NaCl by mass. How many moles of NaCI are in 74 g of this mixture? 9, what is the mass of 2.6 × 1023 molecules of ammonia, NH,? 10. A water solution of sulfuric acid has a density of 1.67 g/mL and is 75 percent H,SO by mass. How many moles of H,SO, are contained in 400. mL of this solution?

Explanation / Answer

Ans. #6. Let the mass of the compound be 100.0 g.

# Mass of C = 51.90 % of 100.0 g = 51.90 g

            Moles of C = Mass /MW = 51.90 g / (12.011 g/ mol) = 4.3210 mol

# Mass of H = 9.80 % of 100.0 g = 9.80 g

            Moles of H = Mass /MW = 9.80 g / (1.00794 g/ mol) = 9.7228 mol

# Mass of C/ = 38.30 % of 100.0 g = 98.30 g

            Moles of Cl = Mass /MW = 38.30 g / (35.4527 g/ mol) = 1.0803 mol

# Simple molar ratio of constituent atoms-   

            C : H : Cl = 4.3210 : 9.7228 : 1.0803 = 3.99 : 9.00 : 1 = 4 : 9 : 1

# Therefore, empirical formula of organic compound = C4H9Cl

#7. Moles of O2 = Mass / MW = 28.4 g / (32.0 g/ mol) = 0.8875 mol

Now,

            No. of O2 molecules = Moles of O2 x Avogadro number

                                                = 0.8875 mol x (6.022 x 1023 molecules/ mol)

                                                = 5.3445 x 1023 molecules

# 1 mol O2 has 2 O-atoms.

So,

            No. of O-atoms = (2 atoms / molecule) x No. of O2 molecules

                                                = 2 x 5.3445 x 1023 molecules

                                                = 1.0690 x 1024 atoms

#8. Mass of NaCl = 48% x 74.0 g = 35.52 g

Moles of NaCl = 35.52 g / (58.44 g/ mol) = 0.6078 mol

#9. Moles of NH3 = No. of molecules / Avogadro number

                                    = 2.6 x 1023 molecules / (6.022 x 1023 molecules/ mol)

                                    = 0.43175 mol

Now,

            Mass of NH3 sample = 0.43175 mol x (17.03056 g/ mol) = 7.353 g

#10. Mass of 400 mL solution = Vol. x Density = 400.0 mL x (1.67 g / mL) = 668.0 g

# Mass of H2SO4 = 75 % x 668.0 g = 501.0 g

Moles of H2SO4 = 501.0 g / (98.07948 g/ mol) = 5.1081 mol

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