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FinnZ [79.3K]
4 years ago
7

What is the empirical formula of ascorbic acid, given that the pseudoformula is C3.407 H4.53 O3.406??

Chemistry
2 answers:
Tom [10]4 years ago
4 0
To determine the empirical formula of ascorbic acid, we divide the lowest fraction with the other values. For this case, we divide everything with 3.406. We do as follows:

C = 3.407 / 3.406 = 1
H = 4.53 / 3.406 = 1.33
O = 3.406 / 3.406 = 1

So, you have CHO

Hope this answers the question. Have a nice day.
almond37 [142]4 years ago
4 0

The empirical formula of ascorbic acid is C₃H₄O₃. Empirical formula is the smallest whole number ratios of elements in a compound.  

<h3>FURTHER EXPLANATION</h3>

To get the empirical formula from mass percent data, the following steps are followed:

  1. Get the mass percent data for all the elements that make up the compound.
  2. Assume that there is 100 grams of sample compound. Determine the equivalent masses of each element using the mass percent given.
  3. Convert the mass of each element to number of moles.
  4. Divide each calculated mole by the smallest mole value.
  5. The quotient will be the subscript of the element in the compound. If a decimal is obtained, round off the answers to the nearest whole number. Some exceptions to this are the following decimals: x.25, x.33, x.50, and x.75. When these decimals are obtained, all the subscripts are multiplied by a number that will result in a whole number.

Applying steps 1 to 3 to the problem leads to the pseudoformula C₃₋₄₀₇H₄₋₅₃O₃₋₄₀₆ where equivalent number of moles are written as the temporary subscripts of the elements in the compound.

<u>STEP 4:</u> Divide the moles by the smallest mole value which in this case is 3.406.

C_{3.407}H_{4.53}O_{3.406}\\\\C_\frac{3.407}{3.406}H_\frac{4.53}{3.406}O_\frac{3.406}{3.406}\\\\\\\boxed {C_1H_{1.33}O_1}

STEP 5: Multiply the subscripts by a number that would give the smallest whole number ratio.  

Since the decimal is 1.33 it cannot be automatically rounded off to 1. It should be multiplied by 3 to get the smallest whole number.

NOTE: All subscripts must be multiplied by the same factor, 3.

3 \times \ C_1H_{1.33}O_1\\\\\boxed {\boxed {C_3H_4O_3}}

<h3>LEARN MORE</h3>
  • Writing Chemical Formula brainly.com/question/4697698
  • Naming Compounds brainly.com/question/8968140
  • Mole Conversion brainly.com/question/12979299

Keywords: empirical formula, chemical formula

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Explanation:

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But in case n- pentane which is a non polar and lack of H-atom attached to electronegative atom. Hence, there is no H- bonding.

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4 years ago
Calculate the molar mass of Fe2(SO4)3. Then Calculate the mass of iron(III) sulfate of 4.05x10^23 formula units. Work must be sh
Talja [164]

Answer:

Molar mass: 399.9 g/mol

The mass of iron(III) sulfate is 269 grams.

Explanation:

First of all, you have to find the molar mass of Fe2(SO4)3.

So you find the molar first for each element

<h3>Fe = 55.8 amu</h3><h3>S = 32.1 amu</h3><h3>O = 16.0 amu</h3>

However, there are 2 iron atoms, 3 sulfur atoms, and 12 oxygen atoms.(Don't forget the 3 outside the parenthesis of SO4).

So you do this

<h3>55.8(2) + 32.1(3) + 16.0(12) = 399.9 </h3>

So the final answer is 399.9 g/mol

Second of all, if you want to find the mass of Fe2(SO4)3, you have to convert from formula units to moles. After that, then you convert from moles to grams.

If you want to convert from molecules to molecules, you have to use Avogadro's number. Avogadro's number is 6.02 x 10^23. Then if you want to convert from moles to grams, you use the molar mass. We already know that the molar mass is 399.9 g/mol.

Therefore, now use dimensional analysis to show your work.

<h3>(4.05 x 10^23) formula units of Fe2(SO4)3 * 1 mol/ 6.02 x 10^23 formula units * 399.9 g/mol / mol</h3>

Formula units in the beginning and the moles will cancel out.

<h3>So [(4.05 x 10^23) / (6.02 x 10^23)] x 399.9 = 269.0357143</h3>

But we need to use significant figures with the least digits(which is 4.05) So we round to the nearest whole number.

<h3>269.0357143 = 269</h3>

So the final answer for the mass of iron(III) sulfate is 269 grams(don't forget the units).

Hope it helped!

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