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zhuklara [117]
3 years ago
10

What is the molecular formula of C3H2N with 260.24g/mol

Chemistry
2 answers:
lisabon 2012 [21]3 years ago
6 0
<span><span>C3</span><span>H2</span>N</span> tells you the lowest possible whole number ratio of atoms in that molecule. That empirical formula must have the same ratio as the actual, molecular formula.
 

The empirical mass and molecular mass must share the same ratio as the formulas do, so find the mass of <span><span>C3</span><span>H2</span>N</span>: <span>3(12g)+2(1g)+14g=52g</span>

Compare that to the molecular weight of 260g.

<span><span><span>MF</span><span>EF</span></span>=<span><span>MW</span><span>EW</span></span></span>so<span><span><span>MF</span><span>EF</span></span>=<span><span>260g</span><span>52g</span></span>=5</span><span>the actual formula is five times heavier than the empirical formula, so take the empirical formula and multiply each subscript by 5. </span>
KATRIN_1 [288]3 years ago
5 0

Answer:

The molecular formula of the compound is  C_{15}H_{10}N_5.

Explanation:

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{molecular mass}}{\text{empirical mass}}

Molecular mass of the compound = 260.24 g/mol

Empirical mass of the compound :

C_3H_2N=3\imes 12 g/mol+3\times 1 g/mol+14 g/mol =52 g/mol

To calculate the valency is :

n=\frac{260.24 g/mol}{52 g/mol}=5

The molecular formula of the compound:

=C_{3\times 5}H_{2\times 5}N_{1\times 5}=C_{15}H_{10}N_5

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4 years ago
A compound is 85.7% C and 14.3% H and has a molar mass of 98g/mol, what is the compounds molecular formula?
vaieri [72.5K]

Answer:

Molecular Formula => (CH₂)₇ => C₇H₁₄      

Explanation:

Empirical ratio is calculated from the sequence...

%/100wt => grams/100wt => moles => mole ratio => reduce mole ratio => Empirical Ratio

C: 85.7% => 85.7g => (85.7/12)mol = 7.14 mole

H: 14.3%  => 14.3g =>     (14.3/1)mol = 14.3 mole

C:H mole ratio => 7.14:14.3

Reduced mole ratio (divide by the smaller mole value) => (7.14/7.14):(14.3/7.14) => Empirical Ratio => 1:2 => Empirical Formula => CH₂

Molecular Wt = Whole No. Multiple of Empirical Formula Wt.

M.Wt = N(Emp Wt) => 98g = N(14g) => N = 7

∴Molecular Formula => (CH₂)₇ => C₇H₁₄

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4 years ago
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