The molar mass and molecular weight of Cocaine (C17H21NO4) is 303.3587 g/mol.
The composition of the C17H21NO4 formula is as follows:
| Element | Symbol | Atomic Weight | Atoms | Total Atomic Weight | Mass Percent |
|---|---|---|---|---|---|
| Oxygen | O | 15.9994 g/mol | 4 | 63.9976 g/mol | 21.0963% |
| Nitrogen | N | 14.0067 g/mol | 1 | 14.0067 g/mol | 4.6172% |
| Hydrogen | H | 1.00797 g/mol | 21 | 21.1674 g/mol | 6.9777% |
| Carbon | C | 12.011 g/mol | 17 | 204.1870 g/mol | 67.3088% |
How to find the molar mass of Cocaine (C17H21NO4)?
The molar mass and molecular weight of Cocaine (C17H21NO4) can be calculated in 4 steps.
Step I: Identify the different elemental atoms present in the C17H21NO4 compound.
The given compound is C17H21NO4. It comprises atoms from 4 different elements i.e., Oxygen (O), Nitrogen (N), Hydrogen (H) and Carbon (C).
Step II: Find the atomic weight of each element in the C17H21NO4.
Here is the list of atomic weights for all elements, let’s check the Oxygen (O), Nitrogen (N), Hydrogen (H) and Carbon (C) atom’s atomic weight.
• The atomic weight of Oxygen (O) is 15.9994 g/mol.
• The atomic weight of Nitrogen (N) is 14.0067 g/mol.
• The atomic weight of Hydrogen (H) is 1.00797 g/mol.
• The atomic weight of Carbon (C) is 12.011 g/mol.
Step III: Determine the number of atoms of each element present in the C17H21NO4 compound.
As per the chemical formula, C17H21NO4, It is made up of 4 Oxygen atoms, 1 Nitrogen atom, 21 Hydrogen atoms and 17 Carbon atoms.
| Element | Number of Atoms |
|---|---|
| O (Oxygen) | 4 |
| N (Nitrogen) | 1 |
| H (Hydrogen) | 21 |
| C (Carbon) | 17 |
Step IV: Calculate the molar mass of the Cocaine (C17H21NO4) compound by applying the formula:
For Cocaine (C17H21NO4):
Substituting into the above formula, the values determined in steps II and III:
Molar mass of C17H21NO4 = [(15.9994 x 4) + (14.0067 x 1) + (1.00797 x 21) + (12.011 x 17)]
Molar mass of C17H21NO4 = 63.9976 + 14.0067 + 21.16737 + 204.186 = 303.3587 g/mol
Result: The molecular weight and molar mass of Cocaine (C17H21NO4) is 303.3587 g/mol.
FAQs
What is the mass percent composition of Oxygen (O) in Cocaine (C17H21NO4)? |
To find the mass percent of Oxygen in C17H21NO4, follow the steps given below:
∴ Mass of Oxygen in C17H21NO4 = 15.9994 x 4 = 63.9976 g/mol. Mass Percent Composition (%) Formula = (Mass of Element in the Compound/Molar Mass of the Compound) x 100 ∴ Mass percent of Oxygen in C17H21NO4 = (Mass of Oxygen in C17H21NO4/Molar mass of C17H21NO4) × 100% Result: C17H21NO4 contains 21.10 % of Oxygen as per its chemical composition.
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What is the mass percent composition of Nitrogen (N) in Cocaine (C17H21NO4)? |
To find the mass percent of Nitrogen in C17H21NO4, follow the steps given below:
∴ Mass of Nitrogen in C17H21NO4 = 14.0067 x 1 = 14.0067 g/mol. Mass Percent Composition (%) Formula = (Mass of Element in the Compound/Molar Mass of the Compound) x 100 ∴ Mass percent of Nitrogen in C17H21NO4 = (Mass of Nitrogen in C17H21NO4/Molar mass of C17H21NO4) × 100% Result: C17H21NO4 contains 4.62 % of Nitrogen as per its chemical composition.
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What is the mass percent composition of Hydrogen (H) in Cocaine (C17H21NO4)? |
To find the mass percent of Hydrogen in C17H21NO4, follow the steps given below:
∴ Mass of Hydrogen in C17H21NO4 = 1.00797 x 21 = 21.1674 g/mol. Mass Percent Composition (%) Formula = (Mass of Element in the Compound/Molar Mass of the Compound) x 100 ∴ Mass percent of Hydrogen in C17H21NO4 = (Mass of Hydrogen in C17H21NO4/Molar mass of C17H21NO4) × 100% Result: C17H21NO4 contains 6.98 % of Hydrogen as per its chemical composition.
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What is the mass percent composition of Carbon (C) in Cocaine (C17H21NO4)? |
To find the mass percent of Carbon in C17H21NO4, follow the steps given below:
∴ Mass of Carbon in C17H21NO4 = 12.011 x 17 = 204.1870 g/mol. Mass Percent Composition (%) Formula = (Mass of Element in the Compound/Molar Mass of the Compound) x 100 ∴ Mass percent of Carbon in C17H21NO4 = (Mass of Carbon in C17H21NO4/Molar mass of C17H21NO4) × 100% Result: C17H21NO4 contains 67.31 % of Carbon as per its chemical composition.
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What is the atomic percentage composition of elements in Cocaine (C17H21NO4)? |
| Atomic percentage composition formula (%) = Number of atoms of a particular element in a compound/Total number of atoms in that compound
For C17H21NO4: The atomic percentage composition of elements in the compound is: Atomic percentage of Oxygen (O) in C17H21NO4: Atomic percentage of Nitrogen (N) in C17H21NO4: Atomic percentage of Hydrogen (H) in C17H21NO4: Atomic percentage of Carbon (C) in C17H21NO4:
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How many grams of Oxygen are present in 1 mole of C17H21NO4? |
| There are 4 O-atoms in a C17H21NO4 molecule. • Atomic weight of a O-atom = 15.9994 g/mol ∴ Total mass of Oxygen in 1 mole of C17H21NO4 = 4 x 15.9994 = 64.00 g. Therefore there are 64.00 grams of Oxygen in 1 mole of C17H21NO4. |
What is the mass of 43 mol of C17H21NO4? |
| The molecular mass of C17H21NO4 is 303.3587 g/mol. This means 303.3587 grams of C17H21NO4 are present per mole. Therefore, we can find the mass of C17H21NO4 in 43 moles as follows: ∴ Moles = Mass/Molar Mass ∴ Mass of C17H21NO4 = Moles x Molar mass ∴ Mass of C17H21NO4 = 43 x 303.3587 = 13044.42 g Thus, the mass of 43 mol of C17H21NO4 is 13044.42 g. |
If you have 76.6711 grams of C17H21NO4, how many moles do you have? |
| ∴ Moles = Mass/Molar mass ∴ Moles of C17H21NO4 = 76.6711/303.3587 = 0.2527 Thus, there are 0.2527 moles of C17H21NO4 in its 76.6711 grams. |
What is the mass in kg of 50.56 x 1025 molecules of C17H21NO4? |
| 1 mole of a substance contains Avogadro number of particles i.e., 6.02 x 1023. Therefore, the number of moles in 50.56 x 1025 molecules of C17H21NO4 are: ∴ Moles of C17H21NO4 = 50.56 x 1025/6.02 x 1023 = 839.9
Now that we have its number of moles, so, we can use the molar mass of C17H21NO4 (303.3587 g/mol) to find its mass as shown below. ∴ Mass of C17H21NO4 = Moles x Molar mass
∴ Mass of C17H21NO4 = 839.9 x 303.3587 = 254779.85 g
Converting mass from grams (g) to kilograms (kg) gives us: ∴ Mass of C17H21NO4 = 254779.85/1000 = 254.78 kg
The mass of 50.56 x 1025 molecules of C17H21NO4 is 254.78 kg.
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What is the molar mass and molecular weight of Cocaine (C17H21NO4)? |
C17H21NO4 is composed of 4 Oxygen (O), 1 Nitrogen (N), 21 Hydrogen (H), and 17 Carbon (C) atoms.
To find the molecular mass of C17H21NO4, one multiplies the atomic weight of each element by its number of atoms in the molecule and then sums the results. ∴ For C17H21NO4, it’s (4 x 15.9994) + (1 x 14.0067) + (21 x 1.0080) + (17 x 12.0110).
Therefore, the molar mass of Cocaine (C17H21NO4) is 303.3587 g/mol.
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About the author
Vishal Goyal is the founder of Topblogtenz, a comprehensive resource for students seeking guidance and support in their chemistry studies. He holds a degree in B.Tech (Chemical Engineering) and has four years of experience as a chemistry tutor. The team at Topblogtenz includes experts like experienced researchers, professors, and educators, with the goal of making complex subjects like chemistry accessible and understandable for all. A passion for sharing knowledge and a love for chemistry and science drives the team behind the website. Let's connect through LinkedIn: https://www.linkedin.com/in/vishal-goyal-2926a122b/