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Sergeeva-Olga [200]

2 years ago

15

penicillin. an important antibiotic (antibacterial agent), was discovered accidentally by the scottish bacteriologist alexander

fleming in 1928, although he was never able to isolate it as a pure compound. this and similar antibiotics have saved millions of lives that would otherwise have been lost to infections. penicillin, like many of the molecules produced by living systems, is a large molecule containing many atoms. one type of penicillin, penicillin f, has the formula c14h20n2so4. Compute the mass percent of each element in this compound.

1 answer:

dmitriy555 [2]2 years ago

7 0

Answer:

mass percent of carbon = 57.78 %

mass percent of hydrogen = 6.40 %

mass percent of nitrogen = 8.96 %

mass percent of oxygen = 20.49 %

mass percent of sulfur = 10.24 %

Explanation:

Given data

Molecular formula = C₁₄H₂₀N₂O₄S

molecular mass (total mass) = 312.39 g/mol

Percentage of carbon = ?

Percentage of hydrogen = ?

Percentage of oxygen = ?

Percentage of nitrogen = ?

Percentage of sulfur = ?

Solution

1st we find out number of moles of each element from the molecular formula

Number of moles of carbon = 14 mol

Number of moles of hydrogen = 20 mol

Number of moles of nitrogen = 2 mol

Number of moles of oxygen = 4 mol

Number of moles of sulfur = 1 mol

Now we find out the mass of each element

as we know that

<em>mass = number of moles × molecular mass</em>

mass of carbon = 14 mol × 12 g/mol

mass of carbon = 168 g

mass of hydrogen = 20 mol × 1 g/mol

mass of hydrogen = 20 g

mass of nitrogen = 2 mol × 14 g/mol

mass of nitrogen = 28 g

mass of oxygen = 4 mol × 16 g/mol

mass of oxygen = 64 g

mass of sulfur = 1 mol × 32 g/mol

mass of sulfur = 32 g

now we find out the mass percent of each element

<em> mass percent = ( mass ÷ total mass ) × 100</em>

mass percent of carbon = ( 168 g ÷ 312.39 g/mol ) × 100

mass percent of carbon = 57.78 %

mass percent of hydrogen = ( 20 g ÷ 312.39 g/mol ) × 100

mass percent of hydrogen = 6.40 %

mass percent of nitrogen = ( 28 g ÷ 312.39 g/mol ) × 100

mass percent of nitrogen = 8.96 %

mass of oxygen =( 64 g ÷ 312.39 g/mol ) × 100

mass percent of oxygen = 20.49 %

mass percent of sulfur = ( 32 g ÷ 312.39 g/mol ) × 100

mass percent of sulfur = 10.24 %

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How many atoms is 3.49x1032 moles of KOH?

antoniya [11.8K]

2.10098*10^47 atoms
Because no. Of atoms = no. Of moles * avogadros no

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A sample of CO2 is collected over water at 23oC. If the total pressure of the sample is 734 torr, and the vapor pressure of wate

noname [10]

Answer:

The partial pressure of CO2 is 712,8 in torr

Explanation:

Molar fraction = Pressure in a compound / Total Pressure

Molar fraction H20 = 21,2 / 734 = 0,0288

Sum of molar fraction in a sample = 1

1 - 0,0288 = 0,9712 (molar fraction of CO2)

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2 years ago

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For the reaction A (g) → 2 B (g), K = 14.7 at 298 K. What is the value of Q for this reaction at 298 K when ∆G = -20.5 kJ/mol?

harina [27]

Answer:

Q= 245 =2.5 * 10^2

Explanation:

ΔG = ΔGº + RTLnQ, so also ΔGº= - RTLnK

R= 8,314 J/molK, T=298K

ΔGº= - RTLnK = - 6659.3 J/mol = - 6.7 KJ/mol

ΔG = ΔGº + RTLnQ → -20.5KJ/mol = - 6.7 KJ/mol + 2.5KJ/mol* LnQ

→ 5.5 = LnQ → Q= 245 =2.5 * 10^2

6 0

2 years ago

Phosphorous acid, h3po3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. calculate the ph f

Varvara68 [4.7K]

Answer:

Explanation:

(a)

Before the addition of KOH :-

Given pKa1 of H3PO3 = 1.30

we know , pKa1 = - log10Ka1

Ka1 = 10-pKa1

Ka1 = 10-1.30

Ka1 = 0.0501

similarly pKa2 = 6.70 ,therefore Ka2 = 1.99 x 10-7

because Ka1 >> Ka2 , therefore pH of diprotic acid i.e H3PO3 can be calculated from first dissociation only .

ICE table is :-

H3PO3 (aq) <-------------> H+ (aq) + H2PO3-(aq)

I 2.4 M 0 M 0 M

C - x + x + x

E (2.4 - x )M x M x M

x = degree of dissociation

Now expression of Ka1 is :

Ka1 = [ H+ ] [ H2PO3-] / [ H3PO3]

0.0501 = x2 / 2.4 - x

on solving for x by using quadratic formula , we have

x = 0.32

Now [ H+ ] = [ H2PO3-] = 0.32 M

pH = - log [H+]

pH = - log 0.32

pH = - ( - 0.495)

pH = 0.495

Hence pH before the addition of KOH = 0.495

(b)

After the addition of 25.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.025 L = 0.06 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

Now 0.06 moles of KOH is equal to the half of the moles required for the first equivalent point . therefore pH at this point is equal to pKa1 .

Hence pH = 1.30 M

(c)

After the addition of 50.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.050 L = 0.12 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

because Number of moles of H3PO4 = Number of moles of KOH

therefore , this point is the first equivalence point

and pH = pKa1 + pKa2 / 2

pH = 1.30 + 6.70 / 2

pH = 4.00

Hence pH = 4.00

(d)

After the addition of 75.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.075 L = 0.18 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

This is the half way of the second equivalence point , therefore pH is equal to pKa2 .

Hence pH = 6.70

5 0

2 years ago

When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr

Amanda [17]

<u>Answer:</u> The value of $K_c$ for the given reaction is 1.435

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of $N_2O_4$ = 9.2 g

Molar mass of $N_2O_4$ = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M$

For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

<u>Initial:</u> 0.20

<u>At eqllm:</u> 0.20-x 2x

We are given:

Equilibrium concentration of $N_2O_4$ = 0.057

Evaluating the value of 'x'

$\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$[NO_2]_{eq}=2x=(2\times 0.143)=0.286M$

$[N_2O_4]_{eq}=0.057M$

Putting values in above expression, we get:

$K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435$

Hence, the value of $K_c$ for the given reaction is 1.435

6 0

2 years ago