Question

calculate the mass percentage of oxygen in a mixture that contains 179 grams of sodium carbonate (na₂co₃) and 149 grams of tartaric acid (c₄h₆o₆). mass percentage o = % use the references to access important values if needed for this question. 3 item attempts remaining try another version submit answer

Explanation:

Step1: Calculate molar mass of Na₂CO₃

The molar mass of Na₂CO₃: $M_{Na_2CO_3}=2\times23 + 12+3\times16=106\ g/mol$. The mass of oxygen in Na₂CO₃ is $m_{O - in - Na_2CO_3}=3\times16 = 48\ g/mol$. The mass of oxygen in 179 g of Na₂CO₃ is $m_{1}=\frac{48}{106}\times179\ g$.

Step2: Calculate molar mass of C₄H₆O₆

The molar mass of C₄H₆O₆: $M_{C_4H_6O_6}=4\times12 + 6\times1+6\times16 = 150\ g/mol$. The mass of oxygen in C₄H₆O₆ is $m_{O - in - C_4H_6O_6}=6\times16=96\ g/mol$. The mass of oxygen in 149 g of C₄H₆O₆ is $m_{2}=\frac{96}{150}\times149\ g$.

Step3: Calculate total mass of the mixture

The total mass of the mixture $m_{total}=179 + 149=328\ g$.

Step4: Calculate total mass of oxygen

$m_{total - O}=m_{1}+m_{2}=\frac{48}{106}\times179+\frac{96}{150}\times149$.
$m_{1}=\frac{48\times179}{106}\approx80.98\ g$, $m_{2}=\frac{96\times149}{150}=95.36\ g$.
$m_{total - O}=80.98 + 95.36=176.34\ g$.

Step5: Calculate mass percentage of oxygen

The mass - percentage of oxygen $=\frac{m_{total - O}}{m_{total}}\times100=\frac{176.34}{328}\times100\approx53.8\%$.

Answer:

53.8%