Question

A. How many atoms of H and O are in 100g of H3PO4? (3 answers)

B. An unknown compound has a mass of 176g/mol and is composed of 40.92% C, 4.58% H and 54.4% O. What is the molecular formula?

C.an unknown compound of CHO has a mass of 0.255g. It undergoes a combustion reaction and yields 0.561g of CO2 and 0.306g of H2O. What is the empirical Formula?

Answer 1

A)

given mass of H3PO4 = 100 g

molar mass of H3PO4 = 97.994 (g/mol)

number moles of H3PO4 = mass / molar mass = 100 g / 97.994 (g/mol) = 1.02 moles

1 mole compound of contains Avagadro number of molecules which is 6.02 x 10^23

there are 3 number of hydrogen atoms present in 1 molecule of H3PO4

so, number of hydrogen atoms present in 1.02 moles = 3 x 1.02 x Avagadro Number

= 3 x 1.02 x  6.02 x 10^23

= 18.42 x 10^23 atoms

there are 4 number of oxygen atoms present in 1 molecule of H3PO4

so, number of hydrogen atoms present in 1.02 moles = 4 x 1.02 x Avagadro Number

= 4 x 1.02 x  6.02 x 10^23

= 24.56 x 10^23 atoms

B)

molar mass of unknown compound = 176 (g/mol)

percentage of carbon (C) present = 40.92 % = 0.4092

molar mass of carbon = 12.01 g

mass of C = 0.4092 x 176 = 72 g

number of carbon atoms present = 72/12 = 6 atoms

percentage of Hydrogen (H) present = 4.58 %

mass of hydrogen = 0.0458 * 176 = 8.06 g

molar mass of H = 1.008

number of atoms of hydrogen present = 8.06/1.008 = 8 atoms

percentage of oxygen (O) present = 54.4 % = 0.544

mass of O present = 0.544* 176 = 95.744

molar mass of oxygen = 15.999

Number of atoms of oxygen present = 95.744 / 15.999 = 6 atoms

molecular formula for unknown compound = C₆H₈O₆

C)

mass of CO2 = 0.561 g

molar mass of CO2 = 44 (g/mol)

Number of moles of CO2 = (0.561/44) = 0.012 moles

mass of H2O = 0.306 g

molar mass of H2O = 18 (g/mol)

Number moles of CO2 = 0.306 / 18 = 0.017 moles

so, number of moles are same for H2O and CO2

so, the empirical formula is CH3CHO