Answer
$$C_2H_3O_3$$
Work Step by Step
1. Calculate the amount of moles of each compound:
$M(CO_2) = 12.0 \space g/mol + 2*16.0 \space g/mol = 44.0 \space g/mol$
$M(H_2O) = 2*1.0 \space g/mol + 16.0 \space g/mol = 18.0 \space g/mol$
$$n(CO_2) = \frac{14.08 \space g}{44.0 \space g/mol} = 0.32 \space mol$$ $$n(H_2O) = \frac{4.32 \space g}{18.0 \space g/mol} = 0.24 \space mol$$
2. Find the amount of C and H moles and their masses:
$$n(C) = n(CO_2) = 0.32 \space mol$$ $$n(H) = n(H_2O) * 2 = 0.24 \space mol * 2 = 0.48 \space mol$$ $$m(C) = 12.0 \space g/mol \times 0.32 \space mol = 3.84 \space g$$ $$m(H) = 1.0 \space g/mol \times 0.48 \space mol = 0.48 \space g$$
3. Find the mass and amount of oxygen atoms:
$$m(Tartaric \space acid) = m(C) + m(H) + m(O)$$ $$m(Tartaric \space acid) -m(C) - m(H) = m(O)$$ $$m(O) = 12.01 \space g - 3.84 \space g - 0.48 \space g = 7.69 \space g$$ $$n(O) = \frac{7.69 \space g}{16.0 \space g/mol} = 0.48 \space mol$$
4. Write the proper formula:
$$C_{0.32}H_{0.48}O_{0.48}$$ $$\div 0.32: C_1H_{1.5}O_{1.5}$$ $$\times 2: C_2H_3O_3$$
