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Atoms and Molecules - Molecular Mass of Covalent Compounds

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The molecular mass of a substance is the sum of the atomic masses of all the atoms in a molecule of the substance. It is therefore the relative mass of a molecule expressed in atomic mass units (uu).

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For covalent compounds, which consist of molecules formed by sharing electrons, we calculate the molecular mass by adding the masses of individual atoms present in one molecule.

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The atomic mass unit (uu) is defined as exactly 112\frac{1}{12}th the mass of one atom of Carbon-12.

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The relative molecular mass is a dimensionless quantity, but when expressed with the unit 'uu', it refers to the mass of a single molecule.

📐Formulae

Molecular Mass=∑(Number of atoms of element×Atomic mass of element)Molecular\ Mass = \sum (Number\ of\ atoms\ of\ element \times Atomic\ mass\ of\ element)

Relative Molecular Mass=Mass of one molecule of the substance112×Mass of one atom of Carbon−12Relative\ Molecular\ Mass = \frac{Mass\ of\ one\ molecule\ of\ the\ substance}{\frac{1}{12} \times Mass\ of\ one\ atom\ of\ Carbon-12}

Mass of AxBy=(x×Atomic mass of A)+(y×Atomic mass of B)Mass\ of\ A_xB_y = (x \times Atomic\ mass\ of\ A) + (y \times Atomic\ mass\ of\ B)

💡Examples

Problem 1:

Calculate the molecular mass of Water (H2OH_2O). (Given atomic masses: H=1uH = 1u, O=16uO = 16u)

Solution:

Molecular mass of H2O=(2×1u)+(1×16u)=2u+16u=18uH_2O = (2 \times 1u) + (1 \times 16u) = 2u + 16u = 18u.

Explanation:

A water molecule contains two atoms of Hydrogen and one atom of Oxygen. We multiply the atomic mass of Hydrogen by 22 and add the atomic mass of Oxygen.

Problem 2:

Calculate the molecular mass of Nitric Acid (HNO3HNO_3). (Given atomic masses: H=1uH = 1u, N=14uN = 14u, O=16uO = 16u)

Solution:

Molecular mass of HNO3HNO_3 is calculated as follows:

  • Mass of 11 atom of H=1×1=1uH = 1 \times 1 = 1u
  • Mass of 11 atom of N=1×14=14uN = 1 \times 14 = 14u
  • Mass of 33 atoms of O=3×16=48uO = 3 \times 16 = 48u

Adding these values: 114+4863\begin{array}{r} 1 \\ 14 \\ + 48 \\ \hline 63 \end{array} Total mass = 63u63u.

Explanation:

We sum the product of the number of atoms and their respective atomic masses for all elements in the formula HNO3HNO_3.

Problem 3:

Calculate the molecular mass of Glucose (C6H12O6C_6H_{12}O_6). (Given atomic masses: C=12uC = 12u, H=1uH = 1u, O=16uO = 16u)

Solution:

Molecular mass of C6H12O6C_6H_{12}O_6:

  • Carbon: 6×12=72u6 \times 12 = 72u
  • Hydrogen: 12×1=12u12 \times 1 = 12u
  • Oxygen: 6×16=96u6 \times 16 = 96u

Total calculation: 7212+96180\begin{array}{r} 72 \\ 12 \\ + 96 \\ \hline 180 \end{array} Total mass = 180u180u.

Explanation:

In a glucose molecule, there are 66 atoms of Carbon, 1212 atoms of Hydrogen, and 66 atoms of Oxygen. Adding their total masses gives 180u180u.