Skip to main content
Chemistry LibreTexts

7.2.3: Practice Empirical and Molecular Formulas

  • Page ID
    236041
  • Percent Composition Problems

    Exercise \(\PageIndex{1}\)

    What is the percent composition of Co(NO3)2?

    Hint: What is formula mass?

    Co mass  \(=1 \times 58.93 = 58.93\) amu

    N mass  \(=2 \times 14.01 = 28.02\) amu

    O mass  \(=6 \times 16.00 = 96.00\) amu

    Total = 182.95 amu

    Answer

    Co mass percent =\(\dfrac{58.93amu}{182.95amu}\times 100\) % = 32.21 %

    N mass percent =\(\dfrac{28.02amu}{182.95amu}\times 100\) % = 15.32 %

    O mass percent =\(\dfrac{96.00amu}{182.95amu}\times 100\) % = 52.47 %

     

    What is the percent composition of Ni2(SO4)3?

    Hint: What is formula mass?

    Ni mass  \(=2 \times 58.69 = 117.38\) amu

    S mass  \(=3 \times 32.07 = 96.21\) amu

    O mass  \(=12 \times 16.00 =192.00\) amu

    Total = 405.59 amu

    Answer

    Ni mass percent =\(\dfrac{117.38amu}{405.59amu}\times 100\) % = 28.94 %

    S mass percent =\(\dfrac{96.21amu}{405.59amu}\times 100\) % = 23.72 %

    O mass percent =\(\dfrac{192.00amu}{405.59amu}\times 100\) % = 47.34 %

     

    You analyze a sample and find that it contains 22.46 g of Fe and 42.76 g Cl.  What is the percent composition of this sample?

    Hint: What is total mass?

    Your sample is 22.46 g + 42.76 g = 65.22 g

    Answer

    Fe mass percent =\(\dfrac{22.46g}{65.22g}\times 100\) % = 34.44 %

    Cl mass percent =\(\dfrac{42.76g}{65.22g}\times 100\) % = 65.56 %

     

     

    Empirical Formula from Percent Composition

    Exercise \(\PageIndex{1}\)

    What is the empirical formula of a compound that is 73.42 % Co and 26.58% O by mass?

    Hint

    Convert 73.42 g Co to moles Co (Co = 58.93 g/mol) and 26.58 g O to moles O (O = 16.00 g/mol).

    Answer

    At first, you get 1.246 mol Co and 1.661 mol O.  Divide both of these by the smaller number (1.246)

    This gives 1 mol Co and 1.33 mol O.  Now multiply both of these by 3.

    This gives 3 mol Co and 4 mol O.

    Co3O4

    What is the empirical formula of a compound that is 26.52 % Cr, 24.53 % S, and 48.95% O by mass?

    Hint

    Convert 26.52 g Cr to moles Cr (Cr = 52.00 g/mol), 24.53 g S to moles S (S = 32.07 g/mol), and 48.95 g O to moles O (O = 16.00 g/mol).

    Answer

    At first, you get 0.5100 mol Cr, 0.7649 mol S, and 3.059 mol O.  Divide all of these by the smallest number (0.5100)

    This gives 1 mol Cr, 1.5 mol S, and 6 mol O.  Now multiply all of these by 2.

    This gives 2 mol Cr, 3 mol S, and 12 mol O.

    Cr2S3O12 (this is actually the compound Cr2(SO4)3)

     

    Empirical and Molecular Formulas

    Exercise \(\PageIndex{1}\)

    For each formula below, give the empirical formula.  Sometimes the formula given is the same as the empirical, sometimes it is different.

    a)  C3H6O3

    b)  N2O4

    c)  Mg3N2

    d)  C7H14O2

    e)  P2O5

    f)  C4H8N2

    Answer a and b

    CH2O and NO2 

    Answer c and d

    Mg3Nand C7H14O2

    Answer e and f

    P2O5 and C2H4

    What is the molecular formula for a compound with empirical formula of CH2O and molecular mass of 150.15 amu?

    Answer

    C5H10O5  (Molecular mass is five times as big as empirical formula mass.)

    What is the molecular formula for a compound with empirical formula of CH2NO2 and molecular mass of 180.12 amu?

    Answer

    C3H6N3O6  (Molecular mass is three times as big as empirical formula mass.)

    What is the molecular formula for a compound with a molecular mass of 86.18 amu that is found to be 83.62 % C and 16.38 % H by mass? (Must find empirical formula first using percent composition.)

    Empirical Formula

    C3H7

    At first, you get 6.963 mol C and 16.22 mol H.  Divide both of these by the smallest number (6.963)

    This gives 1 mol C and 2.33 mol H.  Now multiply both of these by 3.

    This gives 3 mol C and 7 mol H.

    Molecular Formula

    C6H14  (Molecular mass is two times as big as empirical formula mass.)

     

    • Was this article helpful?