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9966 [12]
3 years ago
12

A solid metal oxide crystallizes in a cubic unit cell. In the unit cell there are metal (M) ions on every corner and in the cent

er of every face. Oxide ions (O) are found in half of the tetrahedral holes. What is the empirical formula of this metal oxide?
Chemistry
1 answer:
kotegsom [21]3 years ago
8 0

Answer:

The empirical formula of the solid metal oxide is : MO

Explanation:

M atom is present in the every corner and in the center of every face.

Number of m atoms :

\frac{1}{8}\times 8+\frac{1}{2}\times 6=4

Number of tetrahedral voids in F.C.C = 2n = 2 × 4 = 8

Oxide ion is present in the half of the tetrahedral void

Number oxide ions = \frac{8}{2}=4

The molecular formula of the solid metal oxide is : M_4O_4=MO

The empirical formula represent the lowest number of atoms present in a compound.

The empirical formula of the solid metal oxide is : MO

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The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold.
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Answer:

(a) m_{gold}=7.322g

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(c) \rho _{coin}=15.94g/cm^3

Explanation:

Hello,

(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign  as shown below:

m_{gold}=\frac{m_{tota}*karats}{24}=\frac{7.988g*22}{24}=7.322g

(b) Now, by knowing the density of gold and copper, 19.32 and 8.94 g/cm³ respectively, we compute each volume, by also knowing that the rest of the coin contains copper:

V_{gold}=\frac{m_{gold}}{\rho_{gold}} =\frac{7.322g}{19.32g/cm^3}=0.379cm^3

m_{copper}=7.988g-7.322g=1.09g\\V_{copper}=\frac{m_{copper}}{\rho_{copper}}=\frac{1.09g}{8.94g/cm^3}  \\\\V_{copper}=0.122cm^3

(c) Finally, the volume is computed by dividing the total mass over the total volume containing both gold and copper:

\rho _{coin}=\frac{m_{total}}{V_{gold}+V_{copper}}=\frac{7.988 g}{0.379cm^3+0.122cm^3}\\  \\\rho _{coin}=15.94g/cm^3

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