1. Find the moles of each element obtained from 116 g of substance

Given masses after decomposition:

• Hydrogen $= 7.5\text{ g}$
• Oxygen $= 60\text{ g}$
• Carbon $= 48.5\text{ g}$

Using atomic masses:

• $M_H = 1$
• $M_O = 16$
• $M_C = 12$

So, $n_H = \frac{7.5}{1} = 7.5$ $n_O = \frac{60}{16} = 3.75$ $n_C = \frac{48.5}{12} \approx 4.0417$

2. Find the simplest mole ratio

Divide by the smallest convenient value, $3.75$: $H:O:C = \frac{7.5}{3.75} : \frac{3.75}{3.75} : \frac{4.0417}{3.75}$ $= 2 : 1 : 1.078$

This is very close to: $C:H:O \approx 1:2:1$ which corresponds roughly to an empirical pattern like $CH_2O$.

Because the given masses are in halves ($7.5, 48.5$), this is clearly from approximate analytical data, so we should check the given molecular formulae directly by their mass percentages.

3. Check each option by mass composition

We compare the percentage composition from the data:

For 116 g sample, $\%H = \frac{7.5}{116}\times 100 \approx 6.47\%$ $\%O = \frac{60}{116}\times 100 \approx 51.72\%$ $\%C = \frac{48.5}{116}\times 100 \approx 41.81\%$

Now evaluate each formula.

Option A: $CH_3COOH = C_2H_4O_2$

Molar mass: $2(12)+4(1)+2(16)=24+4+32=60$ Mass percentages: $\%C=\frac{24}{60}\times 100=40\%$ $\%H=\frac{4}{60}\times 100=6.67\%$ $\%O=\frac{32}{60}\times 100=53.33\%$ These are close to the observed values $(41.81\%, 6.47\%, 51.72\%)$. So A agrees.

Option B: $HCHO = CH_2O$

Molar mass: $12+2+16=30$ Mass percentages: $\%C=\frac{12}{30}\times 100=40\%$ $\%H=\frac{2}{30}\times 100=6.67\%$ $\%O=\frac{16}{30}\times 100=53.33\%$ Same percentage composition as option A. So B agrees.

Option C: $CH_3OOCH_3 = C_2H_6O_2$

Molar mass: $24+6+32=62$ Mass percentages: $\%C=\frac{24}{62}\times 100\approx 38.71\%$ $\%H=\frac{6}{62}\times 100\approx 9.68\%$ $\%O=\frac{32}{62}\times 100\approx 51.61\%$ Hydrogen percentage is far too high. So C does not agree.

Option D: $CH_3CHO = C_2H_4O$

Molar mass: $24+4+16=44$ Mass percentages: $\%C=\frac{24}{44}\times 100\approx 54.55\%$ $\%H=\frac{4}{44}\times 100\approx 9.09\%$ $\%O=\frac{16}{44}\times 100\approx 36.36\%$ These do not match. So D does not agree.

4. Final count

The data agrees with:

• $A.\ CH_3COOH$
• $B.\ HCHO$

Hence, the number of formulae that agree is: $\boxed{2}$

5. Comparison with stored answer

Stored correct answer = $2$

Our derived answer also = $2$, so they agree.