Question at position 2 A sample of an oxide of vanadium is separated into its elements, resulting in 1.02 g V and 0.48 g O. Which of the following is the formula of the oxide?V2OV2O5V2O3VO2

Question

Question at position 2 A sample of an oxide of vanadium is separated into its elements, resulting in 1.02 g V and 0.48 g O. Which of the following is the formula of the oxide?V2OV2O5V2O3VO2

BlackTom AI · Accepted Answer

The task is to find the empirical formula of an oxide of vanadium from the masses of V and O obtained after separation.
First, determine the moles of each element from the given masses: vanadium mass = 1.02 g and vanadium atomic weight ≈ 50.94 g/mol, so n(V) = 1.02 / 50.94 ≈ 0.0200 mol. Oxygen mass = 0.48 g and oxygen atomic weight ≈ 16.00 g/mol, so n(O) = 0.48 / 16.00 = 0.0300 mol.
Next, compare the mole ratio by dividing both by the smaller number of moles: 0.0200 mol is the smaller, so n(V)/n(min) ≈ 0.0200 / 0.0200 = 1, and n(O)/n(min) ≈ 0.0300 / 0.0200 = 1.5. A 1:1.5 ratio translates to 2:3 when scaled to whole numbers (multiply both by 2).
Thus the empirical formula is V2O3.
Now evaluate the provided options one by one:
- Option A: V2O. This would correspond to a V:O ratio of 2:1, not matching the calculated 2:3, so it is incorrect.
- Option B: V2O5. This implies a V:O ratio of 2:5, far from 2:3; the oxygen content is too high relative to vanadium, so this is incorrect.
- Option C: V2O3. This matches the derived 2:3 ratio exactly, so this is the correct empirical formula.
- Option D: VO2. This corresponds to a 1:2 ratio, again not matching the calculated 2:3, so it is incorrect.
In summary, the mass data lead to a V:O mole ratio of approximately 1:1.5, which simplifies to 2:3, confirming V2O3 as the correct formula.

Question at position 2 A sample of an oxide of vanadium is separated into its elements, resulting in 1.02 g V and 0.48 g O. Which of the following is the formula of the oxide?V2OV2O5V2O3VO2

查看解析

登录即可查看完整答案

类似问题

A sample of phosphorus of mass 0.808 g reacts with oxygen to form 1.852 g of a phosphorus oxide. Determine the empirical formula of the oxide.

Surgical grade titanium has the mass percent composition of 64.39% titanium, 11.42% vanadium, and 24.19% aluminum. What is the empirical formula of this compound?

A compound that contains only xenon and oxygen is 73.2% Xe by mass. What is the empirical formula for this compound?

A compound has an elemental analysis of 48.2% C, 8.2% H, and 43.2% O by mass. What is the empirical formula?

A compound has an empirical formula of CH and a molecular ion at m/z = 26. The compound could be

A common anti-inflammatory drug is 75.69% C, 8.80% H, and 15.51% O. What is its empirical formula?

A chemist decomposes a compound; the masses of its constituent elements are shown below. Calculate the empirical formula for the compound: 1.21 g Be 2.15 g O 8.63 g S Please report in the form Be#O#S#

A chemist decomposes a compound; the masses of its constituent elements are shown below. Calculate the empirical formula for the compound: 1.01 g Be 1.80 g O 7.19 g S Please report in the form Be#O#S#

更多留学生实用工具

考试浏览器助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单