In an organic compound of molar mass greater than `100` containing only `C`, `H` and `N`, the percentage of `C` is `6` times the percentage of `H` while the sum of the percentage of `C` and H is `1.5` times the percentage of `N`. What is the least molar mass :

Options

A. 175 g/mole
B. 140 g/mole
C. 105 g/mole
D. 210 g/mole

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Define Variables
Let the percentages of carbon (C), hydrogen (H), and nitrogen (N) in the compound be represented as:
- Percentage of C = \( x \)
- Percentage of H = \( y \)
- Percentage of N = \( z \)

### Step 2: Establish Relationships
According to the problem:
1. The percentage of C is 6 times the percentage of H:
\[ x = 6y \]
2. The sum of the percentages of C and H is 1.5 times the percentage of N:
\[ x + y = 1.5z \]

### Step 3: Substitute for y
From the first equation \( x = 6y \), we can express \( y \) in terms of \( x \):
\[ y = \frac{x}{6} \]

### Step 4: Substitute y into the Second Equation
Substituting \( y \) into the second equation:
\[ x + \frac{x}{6} = 1.5z \]
To combine the terms on the left, we can write:
\[ \frac{6x + x}{6} = 1.5z \]
This simplifies to:
\[ \frac{7x}{6} = 1.5z \]

### Step 5: Solve for z
Multiply both sides by 6 to eliminate the fraction:
\[ 7x = 9z \]
Now we can express \( z \) in terms of \( x \):
\[ z = \frac{7x}{9} \]

### Step 6: Find the Total Percentage
Since the total percentage of C, H, and N must equal 100%, we have:
\[ x + y + z = 100 \]
Substituting \( y \) and \( z \) in terms of \( x \):
\[ x + \frac{x}{6} + \frac{7x}{9} = 100 \]

### Step 7: Find a Common Denominator
The common denominator for 6 and 9 is 18. Rewrite the equation:
\[ \frac{18x}{18} + \frac{3x}{18} + \frac{14x}{18} = 100 \]
Combine the terms:
\[ \frac{35x}{18} = 100 \]

### Step 8: Solve for x
Multiply both sides by 18:
\[ 35x = 1800 \]
Now divide by 35:
\[ x = \frac{1800}{35} \approx 51.43 \]

### Step 9: Calculate y and z
Using \( x \) to find \( y \) and \( z \):
\[ y = \frac{x}{6} \approx \frac{51.43}{6} \approx 8.57 \]
\[ z = \frac{7x}{9} \approx \frac{7 \times 51.43}{9} \approx 40 \]

### Step 10: Verify the Percentages
Check if \( x + y + z = 100 \):
\[ 51.43 + 8.57 + 40 = 100 \]
This confirms our calculations are correct.

### Step 11: Calculate Molar Mass
Now, we can find the molar mass using the ratios:
- C: \( x \) = 51.43% → \( \frac{51.43}{100} \times M \) (where \( M \) is the molar mass)
- H: \( y \) = 8.57% → \( \frac{8.57}{100} \times M \)
- N: \( z \) = 40% → \( \frac{40}{100} \times M \)

Assuming a ratio of \( C_aH_bN_c \):
Let \( z = 2 \) (to satisfy the molar mass condition):
- C = 6
- H = 12
- N = 4

Calculating the molar mass:
\[ \text{Molar mass} = 6 \times 12 + 12 \times 1 + 4 \times 14 = 72 + 12 + 56 = 140 \text{ g/mol} \]

### Conclusion
The least molar mass of the compound is **140 g/mol**.