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

Question

BlackTom AI · Accepted Answer

We are given the percent composition of an anti-inflammatory drug: C 75.69%, H 8.80%, O 15.51%. The task is to find the empirical formula.

First, convert each percent to moles by assuming a 100 g sample:
- Carbon: 75.69 g / 12.01 g·mol−1 ≈ 6.30 mol
- Hydrogen: 8.80 g / 1.008 g·mol−1 ≈ 8.73 mol
- Oxygen: 15.51 g / 16.00 g·mol−1 ≈ 0.97 mol
These are the mole ratios, which we then normalize by dividing by the smallest number of moles among the three elements. The smallest is about 0.97 mol (O):
- C: 6.30 / 0.97 ≈ 6.50
- H: 8.73 / 0.97 ≈ 9.00
- O: 0.97 / 0.97 = 1.00
Thus the simplest whole-number ratio is approximately C6.5H9O. Since we generally want whole numbers, multiply by 2 to clear the 0.5: C13H18O2. This gives an empirical formula of roughly C13H18O2 based on the calculated ratios.

Now let's evaluate the provided options in light of this result:
- Option 1: C19H16O4. This greatly differs from the C13H18O2 ratio; the carbon-to-hydrogen ratio and the oxygen count do not match the derived multiples, indicating this is not consistent with the data.
- Option 2: C14H13O4. The hydrogen count is too low relative to carbon and oxygen when compared to the calculated ratio; also the H is far from the expected ~H18, making this unlikely.
- Option 3: C6H9O. This is a very small set of integers and does not reflect the near-doubled values we found after scaling; the C and H counts are roughly a factor of two too small, and it omits one oxygen compared with the derived pattern.
- Option 4: C14H11O2. Here the hydrogen count is significantly lower than the expected ~18, and the carbon count is only slightly higher than C13; this would imply a very different mass distribution than what the percent composition suggests.
- Option 5: C13H18O2. This matches the calculated empirical ratio after scaling and is the closest to the derived C13H18O2, aligning with the mole-ratio approach. Depending on rounding or significant figures in the problem, this is the most consistent choice among the options.

In summary, the process involves converting mass percentages to molar ratios, normalizing by the smallest amount, and scaling to whole numbers. The closest match to the resulting empirical formula after these steps is the set that aligns with roughly C13H18O2, with option 5 (C13H18O2) being the one that best fits the calculated ratios. The remaining options deviate in carbon, hydrogen, and/or oxygen counts in ways that do not correspond to the derived empirical-stoichiometry from the given percentages.

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

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

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

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