Determine the empirical formula of a compound containing 43.3% sodium, 11.3% Carbon, 45.28% Oxygen by mass

N a_{2} C O_{3}

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N a_{2} C_{2} O_{3}

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N a_{2} C O_{4}

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N a_{4} C_{4} O_{6}

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Solution

The correct option is A $N a_{2} C O_{3}$
$\begin{matrix} Element & % & Atomic Mass & % Atomic mass & Simplest atomic Ratio & Simple whole number ratio N a & 43.3 % & 23 & \frac{43.3}{23} = 1.88 & \frac{1.88}{0.94} = 2 & 2 C & 11.3 % & 12 & \frac{11.3}{12} = 0.94 & \frac{0.94}{0.94} = 1 & 1 O & 45.28 % & 16 & \frac{45.28}{16} = 2.83 & \frac{2.83}{0.94} = 3 & 3 \end{matrix}$

$N a_{2} C O_{3}$ is the compound.

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Determine the empirical formula of a compound containing 43.3% sodium, 11.3% Carbon, 45.28% Oxygen by mass

The correct option is A Na2CO3Element%Atomic Mass% Atomic massSimplest atomic RatioSimple whole number ratioNa43.3%2343.323=1.881.880.94=22C11.3%1211.312=0.940.940.94=11O45.28%1645.2816=2.832.830.94=33 Na2CO3 is the compound.