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iragen [17]
3 years ago
15

How do you know how to order an empirical formula?

Chemistry
2 answers:
kolbaska11 [484]3 years ago
8 0

Solving Empirical Formula Problems There are two common types of empirical formula problems. Luckily, the steps to solve either are almost exactly the same. Example #1: Given mass % of elements in a compound.
A compound was found to contain 32.65% Sulfur, 65.3% Oxygen and 2.04% Hydrogen. What is the empirical formula of the compound?
1) The first step in this problem is to change the % to grams.
32.65%→32.65g of S
65.3%→65.3g of O
2.04%→2.04g of H
2) Next divide all the given masses by their molar mass.
32.65g of S/ 32gm-1 = 1.0203 moles of S
65.3g of O/ 16gm-1 = 4.08 moles of O
2.04g of H/ 1.008gm-1 = 2.024 moles of H
3) Then, pick the smallest answer in moles from the previous step and divide all the answers by that. Remember that if you calculate a number that is x0.9 round to the nearest whole number
1.0203 moles of S/ 1.0203 = 1
4.08 moles of O/1.0203 = 3.998 ≈ 4
2.024 moles of H/1.0203 = 1.984 ≈ 2
4) Lastly, the coefficients calculated in the previous step will become the subscripts in the chemical formula.
S = 1
O = 4
H = 2
H2SO4
Example#2: Given the mass of a reactant before a chemical reaction and the mass of a product after a reaction.
When 0.273g of Mg is heated in a Nitrogen (N2) environment a chemical reaction occurs. The product of the reaction is 0.378g . Calculate the empirical formula.
1) In any empirical formula problem you must first find the mass % of the elements in the compound. Since the total mass of the final product was 0.378 we find that:
0.378g total-0.273g magnesium = 0.105g nitrogen
0.105g nitrogen/0.378g total (100) = 27.77%
0.273g magnesium/0.378g total (100) =72.23%
2) Then change the % to grams
27.77%→27.77g of N
72.23%→72.23g of Mg
3) Next, divide all the masses by their respective molar masses.
27.77g/14gm-1 = 1.98 moles N
72.23g/24.31gm-1 = 2.97 moles Mg
inessss [21]3 years ago
3 0

Answer:

Start with the number of grams of each element, given in the problem. the mass of each element = the percent given. Convert the mass of each element to moles using the molar mass from the periodic table. Divide each mole value by the smallest number of moles calculated.

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Answer:

1.) 13 g C₄H₁₀

2.) 41 g CO₂

Explanation:

To find the mass of propane (C₄H₁₀) and carbon dioxide (CO₂), you need to (1) convert mass O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles C₄H₁₀/CO₂ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles C₄H₁₀/CO₂ to mass C₄H₁₀/CO₂ (via molar mass). It is important to arrange the ratios in a way that allows for the cancellation of units. The final answers should have 2 sig figs to match the sig figs of the given value.

Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)

Molar Mass (C₄H₁₀): 58.124 g/mol

Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)

Molar Mass (CO₂): 44.007 g/mol

Molar Mass (O₂): 2(15.998 g/mol)

Molar Mass (O₂): 31.996 g/mol

2 C₄H₁₀ + 13 O₂ ----> 8 CO₂ + 10 H₂O

 48 g O₂             1 mole             2 moles C₄H₁₀            58.124 g
---------------  x  -----------------  x  -------------------------- x  ------------------  =  
                         31.996 g              13 moles O₂               1 mole

=  13 g C₄H₁₀

 48 g O₂             1 mole               8 moles CO₂            44.007 g
---------------  x  -----------------  x  -------------------------- x  ------------------  =  
                         31.996 g              13 moles O₂               1 mole

=  41 g CO₂

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Answer: Antoine Lavoisier.

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