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Relative atomic mass

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Mada za sehemu hiiAtomic StructureMada 5

The Relative Atomic Mass (RAM) of an element is the average mass of all the naturally occurring isotopes of that element, compared to the mass of one atom of carbon-12.

If an element has more than one isotope, its RAM is calculated by considering both the mass and the relative abundance (percentage) of each isotope.

Symbol for relative atomic mass is RAM or Aᵣ

Formula for RAM:

RAM=(A1×P1)+(A2×P2)+...100\text{RAM} = \frac{(A_1 \times P_1) + (A_2 \times P_2) + ...}{100}

Where:

  • A = mass number of the isotope
  • P = percentage abundance of the isotope

Example 1

A sample of chlorine gas contains 75% of isotope ³⁵Cl and 25% of isotope ³⁷Cl.

Find the relative atomic mass (RAM) of chlorine.

Solution:

RAM=(75×35)+(25×37)100=2625+925100=3550100=35.5\text{RAM} = \frac{(75 \times 35) + (25 \times 37)}{100} = \frac{2625 + 925}{100} = \frac{3550}{100} = 35.5

Answer:

The relative atomic mass of chlorine is 35.5

Example 2

A sample of chlorine contains isotopes in the ratio 3:1 (³⁵Cl : ³⁷Cl).

Find the relative atomic mass.

Solution:

Total parts = 3 + 1 = 4

RAM=(3×35)+(1×37)4=105+374=1424=35.5\text{RAM} = \frac{(3 \times 35) + (1 \times 37)}{4} = \frac{105 + 37}{4} = \frac{142}{4} = 35.5

Answer:

RAM = 35.5

Example 3

A sample of oxygen contains isotopes in the ratio 3:2:1 with masses ¹⁶O, ¹⁷O, and ¹⁸O respectively.

Find the relative atomic mass.

Solution:

Total parts = 3 + 2 + 1 = 6

RAM=(3×16)+(2×17)+(1×18)6=48+34+186=100616.7\text{RAM} = \frac{(3 \times 16) + (2 \times 17) + (1 \times 18)}{6} = \frac{48 + 34 + 18}{6} = \frac{100}{6} \approx 16.7

Answer:

RAM of oxygen = 16.7

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