EXPERIMENTAL PROBLEM n. 11 GROUP 7

By: Noora, Lida, Ahsan, Andrea L., Matteo

You receive a sample of a salt formed by an "s-block" metal and an halogen. The amount of this sample contains as many halogen ions as there are in 1,00 grams of sodium chloride. You have to recognize the metal and the halogen in that salt.

Vi viene assegnato un campione di sale formato da un metallo del "blocco s" e da un alogeno. La quantità di questo campione contene tanti ioni alogenuro quanti ve ne sono in 1,00 g di cloruro di sodio. Dovete riconoscere il metallo e l'alogeno nel sale.

Saat suolanäytteen, joka koostuu alkali- tai maa-alkaliryhmän metallista (=s-ryhmä) ja halogeenista. Näyte sisältää yhtä monta halogeeni-ionia kuin 1,00 grammassa natriumkloridia (NaCl) on kloridi-ioneita. Tehtävänä on tunnistaa, mistä metalli- ja halogeeni-ioneista suola koostuu.

1. Useful knowledge (reference data, concepts and principles)

1.1 Colours of alkali metals to the bunsen flame:

Lithium

Sodium

Potassium

Calcium

Strontium

Barium

Deep Purple

Yellow-Orange

Lilac

Orange+Red

Sparkle

Green-Yellow

1.2 Atomic relative mass of sodium and chlorine: Na= 22,99 Cl= 35,33

1.3 Formula Weight of NaCl: 58,44

1.4 Formula of a salt formed by Me⁺ and X^- ions: MeX

1.5 Formula of a salt formed by Me²⁺ and X^- ions:MeX₂

2. Main idea

We recognize the metal with the flame test and the halogen through calculation based on the weight of the salt .

3. Planning

1.weigh the bowl with the salt

2.weigh the empty container

3.kindle the flame

4.we clean nickel wire

5.bathe the wire in the HCl

6 bathe the wire in the wet bowl of salt

7 observe the flame test color since

4. Report

Italian 19-03-2010

Ahsan, Matteo, Andrea L,

Finnish (date of the experiment)

plastic box with salt =3,97 g

empty box =1.89 g

salt = 2,08 g

flame test= green-yellow

empty box=13.5g

box+salt=15.7g

salt=2.2g

flame test= green, yellow

5. Elaboration and Interpretation

Italian

Finnish

salt mass = 3.97-1.89 = 2.08 g

Metal-ion = barium

barium atomic mass = 137.327 amu

Ba ions are Ba²⁺.

our salt formula should be BaCl₂

number of halide ions = number of chloride ions in 1 g of NaCl = number of NaCl couples in 1 g of NaCl = 1 g..................

Salt = 2,2 g

metal-ion= barium

Ba= 137.3
1g :58.44 = 0.0171
0.017 : 2 = 0.0085
0.0085 x 137.3 = 1.16705
2.2 - 1.16705 = 1.03295
1.03295 : 0.0171 = 60.4 > Br
BaBr2

Common conclusions

Questions & Answers

1. Why are atoms emitting light at the flame?

1. Perché gli atomi emettono luce alla fiamma?

Because the atoms are more excitable.

Ripsosta 1.

2. How did you recognize the metal element in the salt?

can be recognized by their color emitted by the flame

3. What measure permitted you to deduce which halogen was in the salt?

3. Quale misura vi ha permesso di dedurre l'alogeno contenuto nel vostro sale?

Answer 3

Risposta 3

4. Does your salt contain more cations or anions?

4. Il vostro sale contiene più cationi o più anioni?

Answer 4

Risposta 4

5. Does your salt contain more mass as halide ions or as metal ions or the same?

5. Il vostro sale contiene una massa maggiore di ioni alogenuroo, di ioni del metallo o la stessa?

Answer 5