# Question #8a2b4

##### 1 Answer

#### Answer:

#### Explanation:

The idea here is that you need to use **Einstein's equation** to convert the **difference in mass** between the helium-4 atom and the **sum** of the hydrogen-3 and hydrogen-1 atoms to **energy**.

#color(blue)(E = m * c^2)" "# , where

*kilograms*

The idea here is that the atomic mass of the helium-4 atom will be **smaller** than the sum of the atomic masses of the hydrogen-3 and hydrogen-1 atoms because some of that mass is converted to *energy*.

The sum of the hydrogen-3 and hydrogen-1 atoms will be

#m_"total" = "3.016049 u" + "1.007825 u" = "4.023874 u"#

The *difference* between this value and the actual mass of the helium-4 atom, also called **mass defect**, will be

#m_"diff" = "4.023874 u" - "4.002603 u" = "0.021271 u"#

Now, the *unified atomic mass unit*,

#"1 u" = 1.660539 * 10^(-27)"kg"#

Use this conversion factor to convert the mass defect from unified atomic mass units to *kilograms*

#0.021271 color(red)(cancel(color(black)("u"))) * (1.660539 * 10^(-27)"kg")/(1color(red)(cancel(color(black)("u")))) = 3.532133 * 10^(-29)"kg"#

Use Einstein's equation to calculate the energy released **per atom** of hydrogen-3 and hydrogen-1

#E = 3.532133 * 10^(-29) * ("299,792,458")^2 overbrace("kg m"^2"s"^(-2))^(color(blue)("Joules"))#

#E = 3.174523* 10^(-12)"J"#

Now, to get the energy released *per gram* of reactant, use **Avogadro's number** to first convert the energy from *Joules per atom* to *Joules per mole*

#3.174523 * 10^(12)"J"/color(red)(cancel(color(black)("atom"))) * (6.022 * 10^(23)color(red)(cancel(color(black)("atoms"))))/"1 mole" = 1.911698 * 10^12"J/mol"#

Now, the *total atomic mass* of the reactants is equal to

#"1 u " = " 1 g/mol"#

to get a total **molar mass** of the reactants of *per gram* of reactant will be

#1.911698 * 10^(12)"J"/color(red)(cancel(color(black)("mol"))) * (1color(red)(cancel(color(black)("mol"))))/"4.023874 g" = color(green)(4.750089 * 10^(11)"J/g")#

The answer is rounded to seven sig figs, the number of sig figs you have for the given atomic masses.