Put the steps below in the order they are performed in the lab experiment by numbering them 1-7. Use a Bunsen burner to heat and dry the hydrate, repeating until the mass no longer decreases. Measure the final masses of the crucible, cover, and anhydrous salt. Compute differences in masses to find the reactants and products. Measure initial masses of the crucible and cover, with and without the hydrate. Use the numbers of moles to compute the mole ratio. Use stoichiometry to convert these masses to moles. Use the mole ratio to write the hydrate’s formula.

1) Measure initial masses of the crucible and cover with and without the hydrate.

2) Use a Bunsen burner to heat and dry the hydrate, repeating until the mass no longer decreases.

3) Measure the final masses of the crucible, cover, and anhydrous salt.

4) Compute differences in masses to find the reactants and products.

5) Use stoichiometry to convert these masses to moles.

6) Use the numbers of moles to compute the mole ratio.

7) Use the mole ratio to write the hydrate’s formula.

Explanation:

A hydrate is any substance that contains water of crystallization. The water of crystallization constitutes part of the mass of the substance. When heated to constant mass, the substance looses its water of crystallization and becomes anhydrous.

The mass of the anhydrous substance is now obtained by weighing again  and subtracted from that of the hydrated substance. The masses are converted to moles using stoichiometry  and the moles are then used to obtain the formula of the hydrate.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol