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elena-s [515]
3 years ago
14

Caffeine, a stimulant found in coffee and soda, has the following elemental composition: 49.48% carbon, 5.19% hydrogen, 16.48% o

xygen and 28.85% nitrogen. The molar mass of caffeine is 119.19 g/mol. Determine the molecular formula of caffeine.
Chemistry
1 answer:
prisoha [69]3 years ago
3 0

Answer:

Explanation:

C = 49.48

H = 5.19

O = 16.48

N = 28.85

ratio of moles

= 49.48 / 12 : 5.19 / 1 : 16.48 / 16 : 28.85 / 14

= 4.123 : 5.19 : 1.03 : 2.06

= 4 : 5 : 1 : 2

so the empirical formula = C₄ H₅O N₂

Let molecular formula = (  C₄ H₅ON₂ )ₙ ,

n ( 48 + 5 + 16 + 28 ) = 119.19

97 n = 194.19

n = 2 ( approx )

molecular formula = C₈ H₁₀O₂ N₄

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What is the reason carbon atoms are the most versatile building blocks of the molecules used by living organisms?
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How Many Moles Of HCl Need To Be Added To 150.0 ML Of 0.50 M NaZ To Have A Solution With A PH Of 6.50
Aleks04 [339]

The number of mole of HCl needed for the solution is 1.035×10¯³ mole

<h3>How to determine the pKa</h3>

We'll begin by calculating the pKa of the solution. This can be obtained as follow:

  • Equilibrium constant (Ka) = 2.3×10¯⁵
  • pKa =?

pKa = –Log Ka

pKa = –Log 2.3×10¯⁵

pKa = 4.64

<h3>How to determine the molarity of HCl </h3>
  • pKa = 4.64
  • pH = 6.5
  • Molarity of salt [NaZ] = 0.5 M
  • Molarity of HCl [HCl] =?

pH = pKa + Log[salt]/[acid]

6.5 = 4.64 + Log[0.5]/[HCl]

Collect like terms

6.5 – 4.64 = Log[0.5]/[HCl]

1.86 = Log[0.5]/[HCl]

Take the anti-log

0.5 / [HCl] = anti-log 1.86

0.5 / [HCl] = 72.44

Cross multiply

0.5 = [HCl] × 72.44

Divide both side by 72.44

[HCl] = 0.5 / 72.4

[HCl] = 0.0069 M

<h3>How to determine the mole of HCl </h3>
  • Molarity of HCl = 0.0069 M
  • Volume = 150 mL = 150 / 1000 = 0.15 L
  • Mole of HCl =?

Mole = Molarity x Volume

Mole of HCl = 0.0069 × 0.15

Mole of HCl = 1.035×10¯³ mole

<h3>Complete question</h3>

How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl

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