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For the atom 166X, find: a) atomic number b) number of neutrons c) number of electrons d) mass number. If another atom is 188X, how is it related to 168X? Find the difference in particles and calculat

Chemistry

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4 parts
a

6\text{6}

b

10\text{10}

c

6\text{6}

d

16\text{16}

16.2

Step-by-step solution

4 steps

For the atom 616X^{16}_{6}X:

Part a

The atomic number of X is the subscript value.

Z=6Z = 6

The atomic number is 6\boxed{\text{6}}.

Part b

10\text{10}

The number of neutrons is the mass number minus the atomic number.

Number of neutrons=A−Z=16−6=10\text{Number of neutrons} = A - Z = 16 - 6 = 10

The number of neutrons is 10\boxed{\text{10}}.

Part c

6\text{6}

For a neutral atom, the number of electrons equals the atomic number.

Number of electrons=Z=6\text{Number of electrons} = Z = 6

The number of electrons is 6\boxed{\text{6}}.

Part d

16\text{16}

The mass number of X is the superscript value.

A=16A = 16

The mass number is 16\boxed{\text{16}}.

If another atom is represented as 818X^{18}_{8}X, its relation to 816X^{16}_{8}X is that they are isotopes.

Difference between 816X^{16}_{8}X and 818X^{18}_{8}X in terms of particles:

  • Both atoms have an atomic number of 8, meaning they both have 8 protons and 8 electrons.
  • They differ in their mass numbers: 816X^{16}_{8}X has a mass number of 16, while 818X^{18}_{8}X has a mass number of 18.
  • This difference in mass number is due to a difference in the number of neutrons.
    • For 816X^{16}_{8}X: Number of neutrons =16−8=8= 16 - 8 = 8.
    • For 818X^{18}_{8}X: Number of neutrons =18−8=10= 18 - 8 = 10.
  • Therefore, 816X^{16}_{8}X has 8 neutrons, and 818X^{18}_{8}X has 10 neutrons.

To show that the relative atomic mass of X would be 16.2, given 90% of 816X^{16}_{8}X and 10% of 818X^{18}_{8}X:

  1. Use the formula for relative atomic mass.

    Relative Atomic Mass (RAM)=(mass of isotope 1×% abundance 1)+(mass of isotope 2×% abundance 2)100\text{Relative Atomic Mass (RAM)} = \frac{(\text{mass of isotope 1} \times \text{\% abundance 1}) + (\text{mass of isotope 2} \times \text{\% abundance 2})}{100}

  2. Substitute the given values.

    RAM=(16×90)+(18×10)100\text{RAM} = \frac{(16 \times 90) + (18 \times 10)}{100}

  3. Calculate the products.

    RAM=1440+180100\text{RAM} = \frac{1440 + 180}{100}

  4. Sum the products and divide by 100.

    RAM=1620100\text{RAM} = \frac{1620}{100} RAM=16.2\text{RAM} = \boxed{16.2}

    The relative atomic mass of X is 16.2.

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