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

`C= 44.12%` % of C

Step-by-step 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= (m_C)/(m_(tot))*100%`

`C= (60g)/(136g)*100%`

`C= 44.12%`