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marissa [1.9K]
2 years ago
6

What is the empirical formula of C6H12O6? C2H4O2 CH2O CH2O2 C2H4O

Chemistry
1 answer:
wariber [46]2 years ago
5 0

Answer: The empirical formula for C6H12O6 is CH2O. Every carbohydrate, be it simple or complex, has an empirical formula CH2O

Explanation:

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SYNTHESIS OFCARBONATECTIONLABORATORY SIMULATIONLab Data- X99.00.10CollectedVolume sodium carbonate (mL)Molarity sodium carbonate
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Answer:

\begin{gathered} \text{Limiting Reagent = Sodium Carbonate} \\ \text{Percent Yield = 98\%} \end{gathered}

Explanation:

The chemical reaction talks about the synthesis of calcium carbonate

It is from the reaction between sodium carbonate and calcium chloride

Let us write the equation of reaction as follows:

Na_2CO_{3(aq)}+CaCl_{2(aq)}\text{ }\rightarrow2NaCl_{(s)\text{ }}+CaCO_{3(aq)}

Firstly, we want to get the expected mass of calcium carbonate

This speaks about getting the theoretical yield based on the equation of reaction

From the data collected, 90 ml of 0.20 M (mol/L) of sodium carbonate gave calcium carbonate

We need to get the actual number of moles of sodium carbonate that reacted

We can get this by multiplying the volume by the molarity (kindly note that we have to convert the volume to Liters by dividing by 1000)

Thus, we have it as:

\frac{90}{1000}\times\text{ 0.1 = 0.009 moles}

Hence, we see that 0.009 moles of sodium carbonate reacted theoretically

Since 1 mole of sodium carbonate gave 1 mole calcium carbonate, it is expected that 0.009 mole of sodium carbonate will give 0.009mole of calcium carbonate

What we have to do now is to get the theoretical grams of calcium carbonate produced

That would be the product of the number of moles of calcium carbonate and its molar mass

The molar mass of calcium carbonate is 100 g/mol

The theoretical yield (expected mass) is thus:

100\text{ g/ mol }\times\text{ 0.009mol = 0.9 g}

Finally, we proceed to get the percentage yield which is calculated using the formula below:

\text{Percent Yield = }\frac{Actual\text{ yield}}{\text{Theoretical yield}}\times\text{ 100 \%}

The actual yield is the observed mass which is given as 0.88 g

The percent yield is thus:

\frac{0.88}{0.9}\times\text{ 100 = }98\text{ \%}

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