Question

What is the empirical formula of vanadium oxide if $2.74g$ of metal oxide contains $1.53g$of metal?
A. ${V_2}{O_3}$
B. $VO$
C. ${V_2}{O_5}$
D. ${V_2}{O_7}$

Answer 1

Hint: The empirical formula is the simplest positive integer ratio of atoms present in a compound. It is calculated by converting the mass of each element to moles using the molar mass from the periodic table and then divides each mole value by the smallest number of moles calculated.

Complete step by step answer:
Given: - $2.74g$ of Metal oxide contains $1.53g$ of metal.
Metal oxide = metal + oxygen
It means metal oxide in $2.74g$ contains = $2.74 - 1.53 = 1.21$ of oxygen.
Amount of metal present in metal oxide = $1.53/2.74 \times 100 = 55.8\% $ .
Amount of oxygen present in metal oxide = $100 - 55.8 = 44.2\% $ .
Number of moles of $V$ = $55.8/51 = 1.09$ ( $51$is a molecular mass) .
Number of moles of $O$ = $44.2/16 = 2.76$ (16 is a molecular mass) .
Ratio for vanadium = $1.09/1.09 = 1$ and ratio of oxygen = $2.76/1.09 = 2.5$ .
The total ratio of metal oxide :
$V = 1 \times 2 = 2$, $O = 2.5 \times 2 = 5$ .
Hence, the empirical formula is ${V_2}{O_5}$ .

So, the correct answer is Option C.

Additional Information:
Vanadium (V) oxide (${V_2}{O_5}$) is commonly known as vanadium pentoxide, an inorganic compound. It is a deep orange color when freshly precipitated from aqueous solution and has a brown/yellow color in solid. It behaves like both amphoteric oxide and an oxidizing agent due to its high oxidation state.

Note:
${V_2}{O_5}$ is used in industrial processes as a catalyst (used to increase the rate of reaction) and helps in the oxidation of ethanol (alcohol) to ethanal (Acetaldehyde) and also helps for the production of oxalic acid, polyamide, phthalic anhydride, and its further products.