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Misha Larkins [42]
3 years ago
11

An unknown substance has the composition of 77.87% c, 11.76% h and 10.37% o. the compound has a molar mass of 154.25 g/mole. wha

t is the molecular formula?
Chemistry
1 answer:
frutty [35]3 years ago
6 0
We need to first calculate the empirical formula. Empirical formula is the simplest ratio of whole numbers of components in a compound,
Mass percentages have been given. We need to then calculate for 100 g of the compound 
                               C                     H                           O
mass                    77.87 g              11.76 g                 <span>10.37 g
number of moles  77.87/12            11.76/1                 10.37/16
moles                  = 6.48                  = 11.76                 =0.648
divide by least number of moles 
                              6.48/0.648       11.76/0.648            0.648/0.648
                              = 10                 =18.1                      = 1
rounded off
C - 10 , H - 18 and O - 1
empirical formula - C</span>₁₀H₁₈O
mass of empirical unit = 12 x 10 + 1x 18 + 16 = 120 + 18 + 16 = 154 
number of empirical units = molecular mass / mass of one empirical unit 
                                         = 154.25 / 154 = 1.00 
Therefore molecular formula = C₁₀H₁₈O
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A 5.00-l tank contains helium gas at 1.50 atm. what is the pressure of the gas in torr if volume is doubled
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3 years ago
How many grams of iron(III) oxide (Fe(OH)3) may theoretically be produced from 85.0 g FeCl3?
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Answer:

56.0 g

Explanation:

Calculation of the moles of FeCl_3 as:-

Mass = 85.0 g

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The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{85.0\ g}{162.2\ g/mol}

Moles= 0.52404\ mol

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FeCl_3+3 NaOH\rightarrow Fe(OH)_3+3 NaCl

1 mole of FeCl_3 on reaction forms 1 mole of Fe(OH)_3

Also,

0.52404 mole of FeCl_3 on reaction forms 0.52404 mole of Fe(OH)_3

Moles of Fe(OH)_3  = 0.52404 moles

Molar mass of Fe(OH)_3  = 106.867 g/mol

Mass = Moles*Molar mass = 0.52404\times 106.867\ g = 56.0 g

8 0
3 years ago
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