Question

Calculate the percent by mass of C in pentaerythritol (C(CH2OH)4)

Answer 1

formula for pentaerythritol = C5H12O4

mass of pentaerythritol = 5 (12) + 12 (1) + 4 (16)
= 60 + 12 + 64
= 138 amu

mass of C = 12 amu


% of C in C5H12O4 = [{5(12)}/ 138 ] × 100
= (60 / 138) × 100 = 43.47%

Answer 2

Answer:

$C= 44.12%$ % of C

Explanation:

Hi, the empirical formula of the pentaerythritol is {tex]C_5H_{12}O_4[/tex]

The molecular weights are:

$M_C=12 g/mol$

$M_H=1 g/mol$

$M_O=16 g/mol$

Due to the empirical formula in 1 mol of pentaerythritol you have 5 mol of C, 12 mol of H and 4 mol O

Taking a calculation base of 1 mol:

$m_C=12 g/mol*5mol$

$m_C=60 g$

$m_H=1 g/mol*12mol$

$m_H=12 g$

$m_O=16 g/mol*4mol$

$m_O=64 g$

The total weight will be:

$m_{tot}=64 g +12 g +60 g= 136 gl$

The C%:

$C= \frac{m_C}{m_{tot}}*100%$

$C= \frac{60g}{136g}*100%$

$C= 44.12%$