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Stella [2.4K]
3 years ago
11

Succinic acid is a substance produced by lichens. Chemical analysis indicates it is composed of 40.68% C, 5.08% H, and 54.24% O.

Determine the empirical formula.
Chemistry
1 answer:
zimovet [89]3 years ago
5 0

Answer:  

C₂H₃O₂

Explanation:  

Assume that you have 100 g of the compound.  

Then you have 40.68 g C, 5.08 g H, and 54.24 g O.  

1. Calculate the <em>moles of each element</em>.

Moles of C  = 40.68 × 1/12.01   = 3.387 mol C

Moles of H =    5.08 × 1/1.008   = 5.010 mol H

Moles of O = 54.24 × 1/16.000 = 3.390 mol O

===============

2. Calculate the <em>molar ratios</em>.  

Divide all values by the smallest number.

C:  3.387/3.387 = 1

H: 5.010/3.387 = 1.479

O: 3.390/3.387 = 1.001

===============

3. <em>Multiply each ratio by 2 </em>

C:  2 × 1       = 2

H: 2 × 1.479 = 2.958

O: 2 × 1.001 = 2.002

===============

4. Determine <em>the empirical formula</em>

Round off all numbers to the closest integer.  

C: 2

H: 3

O: 2

The empirical formula is C₂H₃O₂.  

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<em />

To find limiting reactant, we need to determine the moles of the reactants:

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<em>Moles Cl₂ -Molar mass: 70.9g/mol-:</em>

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8 0
2 years ago
What is the empirical formula of a compound that is 7.74% H and 92.26% C? What is the molecular formula if the molar mass is 78.
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Answer:

For all these questions, we want to find the empirical and molecular formulae of various compounds given their percent composition and molar mass. The technique used to answer one of the questions can accordingly be applied to all of them.

Approaching the first question, we treat the percentages of each element as the mass of that element in a 100 g compound (as the percentages add up to 100%). So, our 100 g compound comprises 7.74 g H and 92.26 g C.

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7.679 mol H/7.679 mol H = 1

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The quotients we calculated represent the subscripts of our compound's empirical formula, which should provide the most simplified whole number ratio of the elements. So the empirical formula of our compound is C₁H₁, or just CH.

Here, it just so happens that we obtained whole number quotients. If we end up with a quotient that isn't a whole number (e.g., 1.5), we would multiply all the quotients by a common number that <em>would </em>give us the most simplified whole number ratio (so, if we had gotten 1 and 1.5, we'd multiply both by 2, and the empirical formula would have subscripts 2 and 3).

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Thus, our molecular formula is C₆H₆.

---

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5. Empirical formula: Cl₄K₂Pt; molecular formula: Cl₄K₂Pt

6. Empirical formula: C₂H₄Cl; molecular formula: C₄H₈Cl₂

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