At conditions 50 K and 20 kPa of temperature and pressure
would helium behave most like an ideal gas. The answer is number 1. This is because
the temperature is so low and the pressure is low too.
The exact molecular mass for butane (C4H10) is
12.0096*4+1.0079*10=58.1174 which is 58.1 to 3 significant figures.
Proportion of carbon in the compound
12.0096*4: 58.1174
=>
48.0384 : 58.1174
The mass of carbon in 2.50 grams of butane can be obtained by proportion, namely
Mass of carbon
= 2.50 * (48.0384/58.1174)
= 2.0664
= 2.07 g (approximated to 3 significant figures)