All categories

2 years ago

How many molecules of ascorbic acid (vitamin c, c6h8o6) are in a 500 mg tablet?

Chemistry

2 answers:

Kamila [148]2 years ago

8 0

1.71 x 10²¹ molecules of ascorbic acid.

<h3>Further explanation</h3>

The Problem:

How many molecules of ascorbic acid (vitamin C or C₆H₈O₆) are in a 500 mg tablet?

The Process:

Step-1

We prepare the molar mass of ascorbic acid or C₆H₈O₆.

Relative atomic mass (in rounding to integers):

C = 12 g/mol
H = 1 g/mol
O = 16 g/mol

In ascorbic acid, there are six carbon atoms, eight hydrogen atoms, and six oxygen atoms. Therefore the molar mass is given by

$\boxed{(12 \times 6) + (1 \times 8) + (16 \times 6) = 176 \ g/moles}$

Step-2

We calculate the moles of 500 mg ascorbic acid. Let us convert it to 0.5 g.

$\boxed{ \ moles = \frac{mass}{molar \ mass} \ } \rightarrow \boxed{ \ moles = \frac{0.5 \ g}{176 \ g/mol} \ }$

Therefore we get 2.84 x 10⁻³ moles.

Step-3

Remember this, one mole contains 6.02 x 10²³ molecules, which is called The Avogadro's number.

Let us convert 2.84 x 10⁻³ moles to the number of molecules (N).

$\boxed{ \ moles = \frac{N}{6.02 \times 10^{23}} \ } \rightarrow \boxed{ \ N = moles \times 6.02 \times 10^{23} \ }$

$\boxed{ \ N = 2.84 \times 10^{-3} \ moles \times 6.02 \times 10^{23} \ \frac{molecules}{mol} \ }$

Thus the amount of molecules in 500 mg C₆H₈O₆ are $\boxed{ \ 1.71 \times 10^{21} \ molecules \ }$ .

- - - - - - -

Quick Steps

$\boxed{ \ \frac{6.02 \times 10^{23} \ molecules}{1 \ mol} \times \frac{1 \ mole}{176 \ gram} \times 0.5 \ grams \ } = \boxed{ \ 1.71 \times 10^{21} \ molecules \ }$

<h3>Learn more </h3>

Determine the mass of aspirin from the number of molecules brainly.com/question/10567477
Determine the number of moles of water produced. brainly.com/question/4142557
What is the formula for the hydrate? brainly.com/question/13012523

Elden [556K]2 years ago

6 0

1) molar mass of the C6H8O6, you need to consult the atomic weight of the C, H and O atoms that are in the periodic table: C is 12; H is 1; O is 16
(12x6)+(1x8)+(16x6)= 176g/mol

176 g = 1 mol
0,5 g = x mol 500mg= 0,5 grams

molar mass = mass÷ moles
176 = 0,5÷ x
x= 2,84 x 10⁻³ mol

2) To find the number of molecules present in those 2,84 x 10⁻³ mol we need to multiple the moles by the Avogadro's constant
No. of molecules = Avogadro's constant x n° of moles
No. of molecules = 6.022 x 10²³ x 2,84 x 10⁻³ 
= 1.71 x 10²¹ molecules of vit C.

You might be interested in

Explain the effects of nh3 and hcl on the cuso4 solution in terms of le chatelier's principle

Fittoniya [83]

The Principle of Le Chatelier states that if a system in equilibrium is subjected to a disturbance, the system will react in such a way that it will diminish the effect of that disturbance. Thus, when the concentration of one of the substances in an equilibrium system is changed, the equilibrium varies in such a way that it can compensate for this change.

For example, if the concentration of one of the reactants is increased, the equilibrium shifts to the right or to the side of the products. Also, if you add more reagents, the reaction will move even more to the right until the balance is re-established again, increasing the quantity of products.

In this way, adding HCl to a solution of CuSO4 will produce the following reaction:

CuSO4 (aq) + 2HCl (aq) ⇔ CuCl2 (aq) + H2SO4 (aq)

Initially the solution of CuSO4 in water will be blue, but when adding HCl the solution will change color to green, since the aqueous solutions of CuCl2 are green. By adding more HCl this color will intensify as the balance shifts to the right, producing more CuCl2 and H2SO4.

On the other hand, adding NH3 to a solution of CuSO4 will produce the following reaction

CuSO4 (aq) + 4NH3 (aq) ⇔ [Cu(NH3)4] SO4 (s)

Thus, by adding NH3 to the CuSO4 solution we will observe the formation of a precipitate corresponding to [Cu(NH3) 4] SO4. When adding more NH3, the formation of more precipitate will be observed as the equilibrium moves to the right, producing a greater quantity of [Cu (NH3) 4] SO4.

6 0

2 years ago

A 20.0-mL sample of lake water was acidified with nitric acid and treated with excess KSCN to form a red complex (KSCN itself is

Novosadov [1.4K]

Answer:

8.09x10⁻⁵M of Fe³⁺

Explanation:

Using Lambert-Beer law, the absorbance of a sample is proportional to its concentration.

In the problem, the Fe³⁺ is reacting with KSCN to produce Fe(SCN)₃ -The red complex-

The concentration of Fe³⁺ in the reference sample is:

4.80x10⁻⁴M Fe³⁺ × (5.0mL / 50.0mL) = 4.80x10⁻⁵M Fe³⁺

Because reference sample was diluted from 5.0mL to 50.0mL.

That means a solution of 4.80x10⁻⁵M Fe³⁺ gives an absorbance of 0.512

Now, as the sample of the lake gives an absorbance of 0.345, its concentration is:

0.345 × (4.80x10⁻⁵M Fe³⁺ / 0.512) = 3.23x10⁻⁵M.

As the solution was diluted from 20.0mL to 50.0mL, the concentration of Fe³⁺ in Jordan lake is:

3.23x10⁻⁵M Fe³⁺ × (50.0mL / 20.0mL) = 8.09x10⁻⁵M of Fe³⁺

4 0

2 years ago

A sample of 0.300 mg pure chromium was added to excess hydrochloric acid to form a 10.0 mL aqueous solution of a chromium (III)

DerKrebs [107]

extinction coefficient (ε) = 347 L·mol⁻¹·cm⁻¹

Explanation:

The chemical reaction between chromium (Cr) and hydrochloric acid (HCl):

2 Cr + 6 HCl → 2 CrCl₃ + 3 H₂

number of moles = mass / molar weight

number of moles of Cr = 0.3 × 10⁻³ (g) / 52 (g/mole)

number of moles of Cr = 5.77 × 10⁻⁶ moles

From the chemical reaction we see that 2 moles of Cr will produce 2 moles of CrCl₃ so 5.77 × 10⁻⁶ moles of Cr will produce 5.77 × 10⁻⁶ moles of CrCl₃.

molar concentration = number of moles / volume (L)

molar concentration of CrCl₃ = 5.77 × 10⁻⁶ / 10 × 10⁻³

molar concentration of CrCl₃ = 5.77 × 10⁻⁴ moles / L

Now we need to transform percent transmittance (%T) in absorbance (A) using the following formula:

A = 2 - log (%T)

A = 2 - log (62.5)

A = 2 - 1.8

A = 0.2

We know that absorbance (A) is defined in respect with extinction coefficient (ε), cell length (l) and concentration (c):

A = εlc

ε = A / lc

ε = 0.2 / (1 × 5.77 × 10⁻⁴)