Identify Z in the following reaction

Caffeine, a stimulant found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 1.000 mg of caffeine

Yuliya22 [10]

Answer:

194 g/mol

Explanation:

1) Content of C:

All the C atoms in the 1.000 mg of caffeine will be found in the 1.813 mg of CO₂.

Mass of C in 1.813 mg of CO₂

Since the atomic mass of C is 12.01 g/mol and the molar of of CO₂ is 44.01, there are 12 mg of C in 44 mg of CO₂ and you can set the proporton:

12.01 mg C / 44.01 mg CO₂ = x / 1.813 g CO₂

⇒ x = 1.813 × 12.01 / 44.01 g of C = 0.49475 mg of C

Number of moles of C

number of moles = mass in g / atomic mass = 0.49475×10⁻³ g / 12.01 g/mol = 4.1195×10⁻⁵ moles = 0.041195 milimol

2) Content of H

All the H atoms in the 1.000 mg of caffeine will be found in the 0.4639 mg of H₂O

Mass of H in 0.4639 mg of H₂O

Since the atomic mass of H is 1.008 g/mol and the molar of of H₂O is 18.015 g/mol, there are 2×1.008 mg of H in 18.015 mg of H₂O and you can set the proporton:

2×1.008 mg H / 18.015 mg H₂O = x / 0.4639 mg H₂O

⇒ x = 0.4639 mg H₂O × 2 × 1.008 mg H / 18.015 mg H₂O = 0.051913 mg H

Number of moles of H

number of moles = mass in g / atomic mass = 0.051913 ×10⁻³ g / 1.008 g/mol = 5.1501× 10⁻⁵ moles = 0.051501 milimol

3) Content of N

All the N atoms in the 1.000 mg of caffeine will be found in the 0.2885 mg of N₂

Mass of N in 0.2885 mg of N₂ is 0.2885 mg

Number of moles of N

number of moles = mass in g / atomic mass = 0.2885 ×10⁻³ g / 14.007 g/mol = 2.0597× 10⁻⁵ moles = 0.020597 milimol

4) Content of O

The mass of O is calculated by difference:

Mass of O = mass of sample - mass of C - mass of H - mass of N

Mass of O = 1.000 mg - 0.49475 mg C - 0.051913 mg H - 0.2885 mg N

Mass of O = 0.1648 mg

Moles of O = 0.1648 × 10 ⁻³ g / 15.999 g/mol = 1.0303×10⁻⁵ mol = 0.01030 milimol

5) Ratios

Divide every number of mililmoles by the smallest number of milimoles:

C: 0.041195 / 0.01030 = 4
H: 0.051501 / 0.01030 = 5
N: 0.020597 / 0.01030 = 2

O: 0.01030 / 0.01030 = 1

C: 4
H: 5
N: 2
O: 1

6) Empirical formula:

C₄H₅N₂O₁

7) Calculate the approximate mass of the empirical formula:

4 × 12 + 5 × 1 + 2 × 14 + 1 × 16 = 97 g/mol

So, since that number is not between 150 and 200 g/mol, multiply by 2: 97 × 2 = 194, which is between 150 and 200.

Thus, the estimate is 194 g/mol

7 0

3 years ago

Which of the following explains the VSEPR geometry of a carbon tetrachloride molecule? (5 points)

Ilya [14]

Answer:

Must be B

Explanation:

I took the test and put A and it was wrong so the answer must be B because i dont think there are lone pairs in this molecule but im not sure

6 0

3 years ago

What is the pH of an aqueous solution with a hydrogen ion concentration of [H ] = 8.2 × 10–4 M?

stira [4]

PH = - log [ H+ ]

pH = - log [ 8.2 x 10⁻⁴]

pH = 3.08

hope this helps!

7 0

3 years ago

26.8 moles of argon contain how many atoms?

stepladder [879]

Answer:

the answers is A

Explanation:

4 0

4 years ago

How many significant figures does 546.000 have

choli [55]

This has only only three sig figs because it is 546.000 the zeros are just place holders so thus meaning they are not significant. It is just another way to write 546 because in that number there is also 3 sig figs. They are just trying to throw you off. Just watch out because if the example is 540 then it is just two significant digits, and it were to be 540. then it be three cause that "." makes the zero significant.

7 0

3 years ago