Question

determine the empirical formula for a compound containing 40% carbon, 6.7% hydrogen and 53.3% oxygen. \\(\bigcirc\\) \\(\ce{cho_{2}}\\) \\(\bigcirc\\) \\(\ce{ch_{2}o}\\) \\(\bigcirc\\) \\(\ce{cho}\\) \\(\bigcirc\\) \\(\ce{c_{3}h_{6}o_{3}}\\) question 25 (0.0862 points) which of the following statements about physical and chemical changes is false \\(\bigcirc\\) phase changes are always physical changes. \\(\bigcirc\\) in a chemical change, matter changes its composition. \\(\bigcirc\\) in a physical change, matter changes its composition. \\(\bigcirc\\) all of these statements are true.

Explanation:

First Question (Empirical Formula)

Step1: Assume 100g of compound

So, mass of C = 40g, H = 6.7g, O = 53.3g.

Step2: Calculate moles

Moles of C: $\frac{40}{12.01} \approx 3.33$; Moles of H: $\frac{6.7}{1.008} \approx 6.65$; Moles of O: $\frac{53.3}{16.00} \approx 3.33$.

Step3: Find mole ratio

Divide by smallest (3.33): C: $\frac{3.33}{3.33}=1$; H: $\frac{6.65}{3.33}\approx 2$; O: $\frac{3.33}{3.33}=1$. So ratio C:H:O = 1:2:1, formula $CH_2O$. Wait, but let's recheck. Wait, 40/12 is ~3.33, 6.7/1 is ~6.7, 53.3/16 is ~3.33. Wait, 6.7/3.33 is ~2. So C:1, H:2, O:1. So empirical formula $CH_2O$. Wait, but the options: $CH_2O$ is an option. Wait, maybe I miscalculated? Wait, 40% C: 40g C, moles C = 40/12 ≈ 3.33. H: 6.7g, moles H = 6.7/1 = 6.7. O: 53.3g, moles O = 53.3/16 ≈ 3.33. Now, divide each by 3.33: C: 1, H: 6.7/3.33 ≈ 2, O: 1. So ratio 1:2:1, so $CH_2O$. Wait, but the selected option was CHO, which is wrong. So correct is $CH_2O$.

Step4: Verify

Empirical formula is simplest whole number ratio. So $CH_2O$ has ratio 1:2:1, which matches the mole ratio.

• Phase changes (like melting) are physical (no composition change) → first statement true.
• Chemical change: matter’s composition changes (new substances) → second true.
• Physical change: matter’s composition does NOT change (e.g., melting ice, still $H_2O$) → third statement is false.
• So the false statement is "In a physical change, matter changes its composition."

Answer:

B. $CH_2O$

Second Question (Physical/Chemical Changes)